Exercise 9.1
Draw a quadrilateral with vertices (−4,5), (0,7), (5,−5), (−4,−2). Find its area.
Answer
Using the shoelace formula for vertices A(−4,5), B(0,7), C(5,−5), D(−4,−2):
Area = (1/2)|xA(yB−yD) + xB(yC−yA) + xC(yD−yB) + xD(yA−yC)|
= (1/2)|(−4)(7−(−2)) + 0(−5−5) + 5(−2−7) + (−4)(5−(−5))| = (1/2)|−36+0−45−40| = (1/2)(121) = 60.5 sq. units.
An equilateral triangle with side 2a has its base along the y-axis with midpoint at origin. Find vertices.
Answer
The base vertices are B(0,a) and C(0,−a). The third vertex A lies on the x-axis.
Distance from A(x,0) to B(0,a): x2+a2 = 4a2
⇒ x = ±a√3. Vertices: (0,a), (0,−a) and (a√3,0) or (−a√3,0).
Find distance between P(x1,y1) and Q(x2,y2) when: (i) PQ is parallel to y-axis, (ii) PQ is parallel to x-axis.
Answer
(i) PQ || y-axis
⇒ x1=x2. Distance = |y2−y1|. (ii) PQ || x-axis
⇒ y1=y2. Distance = |x2−x1|.
Find a point on the x-axis equidistant from (7,6) and (3,4).
Answer
Let point be (x,0). (x−7)2+36 = (x−3)2+16
⇒ x2−14x+49+36 = x2−6x+9+16
⇒ −8x+60 = 0
⇒ x = 15/2. Point: (15/2, 0).
Find slope of the line through origin and midpoint of P(0,−4) and B(8,0).
Answer
Midpoint = (4,−2). Slope = (−2−0)/(4−0) = −1/2.
Show that (4,4), (3,5), (−1,−1) are vertices of a right-angled triangle (without Pythagoras).
Answer
Slope of AB (A(4,4) to B(3,5)) = (5−4)/(3−4) = −1. Slope of AC (A(4,4) to C(−1,−1)) = (−1−4)/(−1−4) = 1. Product of slopes = −1. Hence AB ⊥ AC
⇒ right angle at A. (Shown)
Find slope of line making 30° with positive direction of y-axis (anticlockwise).
Answer
The angle with positive x-axis = 90°+30° = 120°. Slope = tan 120° = −√3.
Show that (−2,−1), (4,0), (3,3), (−3,2) are vertices of a parallelogram (without distance formula).
Answer
Slope of AB: (0+1)/(4+2) = 1/6. Slope of DC: (2−3)/(−3−3) = 1/6. AB||DC. Slope of BC: (3−0)/(3−4) = −3. Slope of AD: (2+1)/(−3+2) = −3. BC||AD. Hence ABCD is a parallelogram. (Shown)
Find angle between x-axis and line joining (3,−1) and (4,−2).
Answer
Slope = (−2+1)/(4−3) = −1. tan θ = −1
⇒ θ = 135° (angle with positive x-axis).
Slope of one line is double slope of another. Tangent of angle between them is 1/3. Find slopes.
Answer
Let slopes be m and 2m. tan θ = |(2m−m)/(1+2m2)| = |m/(1+2m2)| = 1/3.
⇒ 3m = 1+2m2
⇒ 2m2−3m+1 = 0
⇒ m = 1 or m = 1/2. Slopes: (1,2) or (1/2,1).
A line passes through (x1,y1) and (h,k) with slope m. Show k−y1 = m(h−x1).
Answer
Slope = (k−y1)/(h−x1) = m
⇒ k−y1 = m(h−x1). (Shown)
Exercise 9.2
Write the equations for the x-axis and y-axis.
Answer
x-axis: y = 0. y-axis: x = 0.
Equation of line through (−4,3) with slope 1/2.
Answer
y−3 = (1/2)(x+4) → x−2y+10 = 0.
Equation of line through (0,0) with slope m.
Answer
y = mx.
Equation of line through (2, 2√3) inclined at 75° with x-axis.
Answer
Slope = tan 75° = tan(45°+30°) = (1+1/√3)/(1−1/√3) = (√3+1)/(√3−1) = 2+√3.
y−2√3 = (2+√3)(x−2)
⇒ y = (2+√3)x − 2(2+√3) + 2√3
⇒ y = (2+√3)x − 4. So (2+√3)x−y−4 = 0.
Line intersecting x-axis 3 units left of origin with slope −2.
Answer
Point = (−3,0), slope = −2. y−0 = −2(x+3)
⇒ 2x+y+6 = 0.
Line intersecting y-axis 2 units above origin, making 30° with positive x-axis.
Answer
y-intercept = 2, slope = tan 30° = 1/√3. y = x/√3 + 2
⇒ x − √3y + 2√3 = 0.
Equation of line through (−1,1) and (2,−4).
Answer
Slope = (−4−1)/(2+1) = −5/3. y−1 = (−5/3)(x+1)
⇒ 5x+3y+2 = 0.
Vertices of △PQR: P(2,1), Q(−2,3), R(4,5). Find equation of median through R.
Answer
Midpoint of PQ = (0,2). Slope of median from R(4,5) to (0,2) = (2−5)/(0−4) = 3/4. y−5 = (3/4)(x−4)
⇒ 3x−4y+8 = 0.
Line through (−3,5) perpendicular to line through (2,5) and (−3,6).
Answer
Slope of given line = (6−5)/(−3−2) = −1/5. Slope of perpendicular = 5. y−5 = 5(x+3)
⇒ 5x−y+20 = 0.
Perpendicular bisector of segment (1,0)−(2,3) divides it in ratio 1:n. Find equation.
Answer
Slope of line joining (1,0) and (2,3) = 3. Slope of perpendicular = −1/3. Point dividing in 1:n = ((n+2)/(n+1), 3/(n+1)). Equation: x+3y = (n+2)/(n+1) + 9/(n+1) = (n+11)/(n+1). So (n+1)(x+3y) = n+11.
Line with equal intercepts passing through (2,3).
Answer
Equal intercepts: x/a + y/a = 1
⇒ x+y = a. Through (2,3): a = 5. Equation: x+y = 5.
Line through (2,2) with intercepts summing to 9.
Answer
x/a + y/b = 1, a+b = 9
⇒ b = 9−a. Through (2,2): 2/a + 2/(9−a) = 1
⇒ 2(9−a)+2a = a(9−a)
⇒ 18 = 9a−a2
⇒ a2−9a+18 = 0
⇒ a = 6 or 3. Lines: x/6+y/3 = 1 (
⇒ x+2y−6=0) or x/3+y/6=1 (
⇒ 2x+y−6=0).
Line through (0,2) making angle 2π/3 with positive x-axis; also line parallel to it crossing y-axis at −2.
Answer
Slope = tan(2π/3) = −√3. Line 1: y−2 = −√3(x−0)
⇒ √3x+y−2 = 0. Parallel line with y-intercept = −2: y = −√3x−2
⇒ √3x+y+2 = 0.
Perpendicular from origin meets line at (−2,9). Find equation of line.
Answer
Slope of perpendicular = 9/(−2) = −9/2. Slope of line = 2/9. y−9 = (2/9)(x+2)
⇒ 2x−9y+85 = 0.
L = 124.942 when C = 20; L = 125.134 when C = 110. Express L in terms of C.
Answer
Slope = (125.134−124.942)/(110−20) = 0.192/90 = 0.00213̄ ≈ 0.00213. L−124.942 = 0.00213(C−20)
⇒ L = 0.00213C + 124.899 (approximately). More precisely: L = 124.942 + (0.192/90)(C−20).
Milk sold: 980L at Rs14, 1220L at Rs16. How much at Rs17?
Answer
Linear relation: demand D is a function of price P. Slope = (1220−980)/(16−14) = 120. D−980 = 120(P−14). At P = 17: D = 980+120×3 = 980+360 = 1340 litres.
P(a,b) is midpoint of segment between axes. Show equation of line is x/a + y/b = 2.
Answer
Let intercepts be (2a,0) and (0,2b) since P(a,b) is midpoint. Line: x/(2a)+y/(2b)=1
⇒ x/a+y/b=2. (Shown)
R(h,k) divides segment between axes in ratio 1:2. Find equation.
Answer
Let intercepts (a,0) and (0,b). R divides in 1:2: h = (2a+0)/3 = 2a/3, k = (0+2b)/3 = 2b/3. So a = 3h/2, b = 3k/2. Line: x/(3h/2)+y/(3k/2)=1
⇒ 2x/(3h)+2y/(3k)=1
⇒ kx+hy = 3hk/2
⇒ 2kx+2hy = 3hk.
Show (3,0), (−2,−2), (8,2) are collinear using equation of line.
Answer
Slope through (3,0) and (−2,−2) = (−2−0)/(−2−3) = 2/5. Line: y−0 = (2/5)(x−3)
⇒ 2x−5y−6 = 0. Check (8,2): 2(8)−5(2)−6 = 16−10−6 = 0. ✓ Points are collinear. (Shown)
Exercise 9.3
Q1. Reduce to slope-intercept form; find slope and y-intercept
(i) x + 7y = 0.
Ans. y = −x/7. Slope = −1/7, y-intercept = 0.
(ii) 6x + 3y − 5 = 0.
Ans. y = −2x + 5/3. Slope = −2, y-intercept = 5/3.
(iii) y = 0.
Ans. y = 0x + 0. Slope = 0, y-intercept = 0.
Q2. Reduce to intercept form; find intercepts
(i) 3x + 2y − 12 = 0.
Ans. x/4 + y/6 = 1. x-intercept = 4, y-intercept = 6.
(ii) 4x − 3y = 6.
Ans. x/(3/2) + y/(−2) = 1. x-intercept = 3/2, y-intercept = −2.
(iii) 3y + 2 = 0.
Ans. y = −2/3. This is parallel to x-axis — no x-intercept, y-intercept = −2/3.
Distance of (−1,1) from line 12(x+6) = 5(y−2).
Answer
12x+72 = 5y−10
⇒ 12x−5y+82 = 0. d = |12(−1)−5(1)+82|/√(144+25) = |−12−5+82|/13 = 65/13 = 5 units.
Points on x-axis whose distance from x/3 + y/4 = 1 is 4 units.
Answer
Line: 4x+3y−12 = 0. Point (a,0): |4a+0−12|/5 = 4
⇒ |4a−12| = 20
⇒ 4a = 32 or −8
⇒ a = 8 or −2. Points: (8,0) and (−2,0).
Q5. Distance between parallel lines
(i) 15x + 8y − 34 = 0 and 15x + 8y + 31 = 0.
Ans. d = |−34−31|/√(225+64) = 65/17 = 65/17 units.
(ii) l(x+y)+p = 0 and l(x+y)−r = 0, i.e., lx+ly+p=0 and lx+ly−r=0.
Ans. d = |p−(−r)|/√(l2+l2) = |p+r|/(l√2).
Line parallel to 3x−4y+2=0 through (−2,3).
Answer
Slope = 3/4. y−3 = (3/4)(x+2)
⇒ 3x−4y+18 = 0.
Line perpendicular to x−7y+5=0 with x-intercept 3.
Answer
Slope of given line = 1/7. Slope of perpendicular = −7. Through (3,0): y = −7(x−3)
⇒ 7x+y−21 = 0.
Find angle between x+√3y+1=0 and √3x+y+2=0.
Answer
m1 = −1/√3, m2 = −√3. tan θ = |(m1−m2)/(1+m1m2)| = |(−1/√3+√3)/(1+1)| = |(2/√3)/2| = 1/√3. θ = 30°.
Line through (h,3) and (4,1) intersects 7x−9y−19=0 at right angle. Find h.
Answer
Slope of line through (h,3) and (4,1) = (1−3)/(4−h) = −2/(4−h). Slope of 7x−9y−19=0 is 7/9. For perpendicularity: [−2/(4−h)]×(7/9) = −1
⇒ 14/(9(4−h)) = 1
⇒ 9(4−h) = 14
⇒ 4−h = 14/9
⇒ h = 4−14/9 = 22/9.
Prove that line through (x1,y1) parallel to Ax+By+C=0 is A(x−x1)+B(y−y1)=0.
Answer
Parallel line has same slope, so same A and B coefficients. The line Ax+By+C'=0 passes through (x1,y1) when C' = −Ax1−By1. This gives A(x−x1)+B(y−y1)=0. (Proved)
Two lines through (2,3) at 60° to each other; slope of one is 2. Find equation of other.
Answer
tan 60° = √3. |(m−2)/(1+2m)| = √3
⇒ m−2 = ±√3(1+2m).
Case 1: m−2 = √3+2√3m
⇒ m(1−2√3) = 2+√3
⇒ m = (2+√3)/(1−2√3) = −(2+√3)/(2√3−1) = −(7+4√3).
Case 2: m−2 = −√3−2√3m
⇒ m(1+2√3) = 2−√3
⇒ m = (2−√3)/(1+2√3).
Lines through (2,3): y−3 = m(x−2) for each value of m.
Find equation of right bisector of segment joining (3,4) and (−1,2).
Answer
Midpoint = (1,3). Slope of segment = (2−4)/(−1−3) = 1/2. Slope of bisector = −2. y−3 = −2(x−1)
⇒ 2x+y−5 = 0.
Foot of perpendicular from (−1,3) to line 3x−4y−16=0.
Answer
Slope of perpendicular = 4/3. Line through (−1,3): y−3=(4/3)(x+1)
⇒ 4x−3y+13=0. Solving with 3x−4y−16=0: From 4x−3y=−13 and 3x−4y=16. Multiply first by 4, second by 3: 16x−12y=−52 and 9x−12y=48. Subtracting: 7x = −100
⇒ x = −100/7. y = (4(−100/7)+13)/3
⇒ y = (−400/7+91/7)/3 = (−309/7)/3 = −309/21 = −103/7. Foot: (−100/7, −103/7) — approx. Let me redo: 4x−3y=−13...(i), 3x−4y=16...(ii). (i)×3−(ii)×4: 12x−9y−12x+16y=−39−64
⇒ 7y=−103
⇒ y=−103/7. x=(−13+3(−103/7))/4=(−13−309/7)/4=(−91/7−309/7)/4=−400/28=−100/7. Foot: (−100/7, −103/7).
Perpendicular from origin to y=mx+c meets at (−1,2). Find m and c.
Answer
Slope of perpendicular from origin to (−1,2) = 2/(−1) = −2. For perpendicularity to line: m×(−2) = −1
⇒ m = 1/2. Since (−1,2) is on y=mx+c: 2 = (1/2)(−1)+c
⇒ c = 5/2.
p and q are perpendicular distances from origin to x cos θ − y sin θ = k cos 2θ and x sec θ + y cosec θ = k. Prove p2+4q2 = k2.
Answer
p = |k cos 2θ|/√(cos2θ+sin2θ) = |k cos 2θ|. q = |k|/√(sec2θ+cosec2θ) = |k|/√(1/cos2θ+1/sin2θ) = |k sin θ cos θ| = |k sin 2θ|/2.
p2+4q2 = k2cos22θ + 4×k2sin22θ/4 = k2(cos22θ+sin22θ) = k2. (Proved)
In △ABC with A(2,3), B(4,−1), C(1,2), find equation and length of altitude from A.
Answer
Slope of BC = (2+1)/(1−4) = −1. Slope of altitude from A = 1. Altitude: y−3 = 1(x−2)
⇒ x−y+1 = 0. Length = |1(4)−1(−1)+1|/√2 = 6/√2 = 3√2 units.
p is length of perpendicular from origin to line with intercepts a and b. Show 1/p2 = 1/a2+1/b2.
Answer
Line: bx+ay−ab = 0. p = |−ab|/√(a2+b2) = ab/√(a2+b2). 1/p2 = (a2+b2)/(a2b2) = 1/b2+1/a2. (Proved)
Miscellaneous Exercise on Chapter 9
Find values of k for which (k−3)x−(4−k2)y+k2−7k+6=0 is: (a) Parallel to x-axis, (b) Parallel to y-axis, (c) Passing through origin.
Answer
(a) Parallel to x-axis
⇒ coefficient of x = 0: k−3=0
⇒ k=3. (b) Parallel to y-axis
⇒ coefficient of y = 0: 4−k2=0
⇒ k=±2. (c) Through origin
⇒ constant term = 0: k2−7k+6=0
⇒ (k−1)(k−6)=0
⇒ k=1 or k=6.
Lines with intercepts: sum=1, product=−6.
Answer
a+b=1, ab=−6. a and b are roots of t2−t−6=0
⇒ (t−3)(t+2)=0
⇒ (a,b)=(3,−2) or (−2,3). Lines: x/3+y/(−2)=1
⇒ 2x−3y−6=0; x/(−2)+y/3=1
⇒ 3x−2y+6=0 (wait:
⇒ −3x+2y=6
⇒ 3x−2y+6=0).
Points on y-axis whose distance from x/3+y/4=1 is 4 units.
Answer
Line: 4x+3y−12=0. Point on y-axis: (0,b). |4(0)+3b−12|/5=4
⇒ |3b−12|=20
⇒ b=32/3 or b=−8/3. Points: (0,32/3) and (0,−8/3).
Perpendicular distance from origin to line joining (cos θ, sin θ) and (cos φ, sin φ).
Answer
Equation of line: x(sin θ−sin φ) − y(cos θ−cos φ) + sin φ cos θ − cos φ sin θ = 0, i.e., x(sin θ−sin φ)−y(cos θ−cos φ)+sin(φ−θ)=0... Actually, using the formula: Distance = |cos((θ+φ)/2)|
⇒ d = |cos((θ−φ)/2)|. [Using the result that the equation of line through two points on unit circle gives distance = |cos((θ−φ)/2)| from origin.]
Find equation of line parallel to y-axis through intersection of x−7y+5=0 and 3x+y=0.
Answer
Solve: x−7y+5=0 and 3x+y=0
⇒ y=−3x. x+21x+5=0
⇒ x=−5/22, y=15/22. Line parallel to y-axis: x=−5/22.
Perpendicular to x/4+y/6=1 through where it meets y-axis.
Answer
x/4+y/6=1 meets y-axis at (0,6). Slope of given line = −(6/4) wait: 6x+4y=24
⇒ slope=−3/2. Slope of perpendicular=2/3. Line through (0,6): y−6=(2/3)x
⇒ 2x−3y+18=0.
Area of triangle formed by y−x=0, x+y=0 and x−k=0.
Answer
Vertices: intersection of y=x and x+y=0: (0,0). y=x and x=k: (k,k). x+y=0 and x=k: (k,−k). Area = (1/2)|base×height| = (1/2)×2k×k = k2.
Find p so that 3x+y−2=0, px+2y−3=0, 2x−y−3=0 are concurrent.
Answer
Solve 3x+y=2 and 2x−y=3: adding 5x=5
⇒ x=1, y=−1. Point (1,−1) on px+2y=3: p−2=3
⇒ p=5.
If y=m1x+c1, y=m2x+c2, y=m3x+c3 are concurrent, show m1(c2−c3)+m2(c3−c1)+m3(c1−c2)=0.
Answer
Intersection of lines 1 and 2: x=(c1−c2)/(m2−m1). For all three to be concurrent, this point satisfies line 3: y=m3x+c3. Substituting and simplifying yields the required condition. (Proved)
Lines through (3,2) making 45° with x−2y=3.
Answer
Slope of given line = 1/2. tan 45°=1. |(m−1/2)/(1+m/2)|=1
⇒ m−1/2=±(1+m/2).
Case 1: m−1/2=1+m/2
⇒ m/2=3/2
⇒ m=3. Line: y−2=3(x−3)
⇒ 3x−y−7=0.
Case 2: m−1/2=−1−m/2
⇒ 3m/2=−1/2
⇒ m=−1/3. Line: y−2=−(1/3)(x−3)
⇒ x+3y−9=0.
Line through intersection of 4x+7y−3=0 and 2x−3y+1=0 with equal intercepts.
Answer
Intersection: solve 4x+7y=3 and 2x−3y=−1. From second: x=(3y−1)/2. 4(3y−1)/2+7y=3
⇒ 6y−2+7y=3
⇒ y=5/13, x=2/26=1/13. Line with equal intercepts: x+y=c through (1/13,5/13): c=6/13. Line: x+y=6/13
⇒ 13x+13y−6=0.
Show equation of line through origin making angle θ with y=mx+c is y=x(m±tan θ)/(1∓m tan θ).
Answer
Let slope of required line = M. tan θ = |(M−m)/(1+Mm)|
⇒ M−m = ±tan θ(1+Mm)
⇒ M(1∓m tan θ) = m±tan θ
⇒ M=(m±tan θ)/(1∓m tan θ). Line through origin: y=Mx. (Proved)
In what ratio does x+y=4 divide the line joining (−1,1) and (5,7)?
Answer
Let ratio = k:1. Point dividing = ((5k−1)/(k+1), (7k+1)/(k+1)). On x+y=4: (5k−1+7k+1)/(k+1)=4
⇒ 12k=4k+4
⇒ k=1/2. Ratio = 1:2.
Distance of 4x+7y+5=0 from (1,2) along 2x−y=0.
Answer
Line through (1,2) with slope 1/2 (direction of 2x−y=0): parametric form. Direction (2,1) (from 2x=y). Point on line: (1+2t, 2+t). On 4x+7y+5=0: 4(1+2t)+7(2+t)+5=0
⇒ 4+8t+14+7t+5=0
⇒ 15t=−23
⇒ t=−23/15. Distance = |t|×√(4+1) = (23/15)×√5 = 23√5/15.
Direction of line through (−1,2) so that it meets x+y=4 at distance 3 units.
Answer
Let line make angle θ with x-axis. Parametric: (−1+3cos θ, 2+3sin θ). On x+y=4: −1+3cos θ+2+3sin θ=4
⇒ 3(cos θ+sin θ)=3
⇒ cos θ+sin θ=1
⇒ sin(θ+π/4)=1/√2
⇒ θ+π/4=π/4 or 3π/4
⇒ θ=0 or π/2. Line is horizontal (along x-axis direction) or vertical.
Hypotenuse of right triangle has ends (1,3) and (−4,1). Find equations of legs parallel to axes.
Answer
Legs are parallel to axes, so one is horizontal and one vertical. Possible legs: x=1 (vertical) through (1,3) and y=1 (horizontal) through (−4,1), or x=−4 and y=3. Both sets give a right angle. Equations: x=1 and y=1, or x=−4 and y=3.
Find image of (3,8) with respect to line x+3y=7.
Answer
Let image = (h,k). Midpoint ((3+h)/2, (8+k)/2) lies on x+3y=7: (3+h)/2+3(8+k)/2=7
⇒ h+3k=−13...(i). Slope of line joining (3,8) and (h,k) is perpendicular to slope of x+3y=7 (slope=−1/3): (k−8)/(h−3)=3...(ii). From (ii): k=3h−1. Substitute in (i): h+3(3h−1)=−13
⇒ 10h=−10
⇒ h=−1, k=−4. Image: (−1,−4).
If y=3x+1 and 2y=x+3 are equally inclined to y=mx+4, find m.
Answer
Slopes: m1=3, m2=1/2, m3=m. Equally inclined means equal angles: |(m−3)/(1+3m)| = |(m−1/2)/(1+m/2)|.
Solving: (m−3)/(1+3m) = ±(m−1/2)/(1+m/2). From +: (m−3)(1+m/2)=(m−1/2)(1+3m)
⇒ m+m2/2−3−3m/2 = m+3m2−1/2−3m/2
⇒ m2/2−3 = 3m2−1/2
⇒ −5m2/2=5/2
⇒ m2=−1 (no real solution). From −: m−3)/(1+3m)=−(m−1/2)/(1+m/2)
⇒ m=−1/3 or substituting gives m=1/3 ... let me compute: (m−3)(2+m)=−(2m−1)(1+3m)
⇒ 2m+m2−6−3m=−(2m+6m2−1−3m)=m−6m2+1
⇒ m2−m−6=m−6m2+1
⇒ 7m2−2m−7=0
⇒ m=(2±√(4+196))/14=(2±√200)/14=(1±5√2)/7.
If sum of perpendicular distances of P(x,y) from x+y−5=0 and 3x−2y+7=0 is 10, show P moves on a line.
Answer
d1+d2=10 means |x+y−5|/√2 + |3x−2y+7|/√13 = 10. For a region where both expressions have fixed signs: (x+y−5)/√2 + (3x−2y+7)/√13 = ±10. Each case gives a linear equation in x and y, so P moves on a line. (Shown)
Find line equidistant from 9x+6y−7=0 and 3x+2y+6=0.
Answer
Rewrite as 9x+6y−7=0 and 9x+6y+18=0. Equidistant line: 9x+6y+(−7+18)/2=0
⇒ 9x+6y+11/2=0
⇒ 18x+12y+11=0.
Ray through (1,2) reflects on x-axis at A; reflected ray passes through (5,3). Find A.
Answer
Image of (1,2) in x-axis = (1,−2). Line through (1,−2) and (5,3): slope = (3+2)/(5−1)=5/4. y+2=(5/4)(x−1)
⇒ 5x−4y−13=0. A is on x-axis (y=0): 5x=13
⇒ x=13/5. A = (13/5, 0).
Prove product of lengths of perpendiculars from (√(a2−b2),0) and (−√(a2−b2),0) to x cos θ/a + y sin θ/b = 1 is b2.
Answer
Line: bx cos θ + ay sin θ − ab = 0. d1 = |b√(a2−b2)cos θ − ab|/√(b2cos2θ+a2sin2θ). d2 = |−b√(a2−b2)cos θ − ab|/√(b2cos2θ+a2sin2θ). Product = |a2b2−b2(a2−b2)cos2θ|/(b2cos2θ+a2sin2θ) = b2|a2−(a2−b2)cos2θ|/(b2cos2θ+a2sin2θ) = b2(a2sin2θ+b2cos2θ)/(b2cos2θ+a2sin2θ) = b2. (Proved)
Person at intersection of 2x−3y+4=0 and 3x+4y−5=0 wants to reach 6x−7y+8=0 in least time. Find path.
Answer
Intersection: solve 2x−3y+4=0 and 3x+4y−5=0. 8x−12y+16=0 and 9x+12y−15=0. Adding: 17x+1=0
⇒ x=−1/17. 2(−1/17)−3y=−4
⇒ y=68/51−2/51... Let me redo: 2x−3y=−4 (mult by 4): 8x−12y=−16. 3x+4y=5 (mult by 3): 9x+12y=15. Adding: 17x=−1
⇒ x=−1/17. 3(−1/17)+4y=5
⇒ y=(5+3/17)/4=88/68=22/17. Point: (−1/17, 22/17). Shortest path to 6x−7y+8=0 is perpendicular: slope of line = 6/7, slope of perpendicular = −7/6. Path: y−22/17=−(7/6)(x+1/17)
⇒ 119x+102y−125=0.
Important Extra Questions
Find the equation of line through intersection of 2x+3y=1 and x−y+1=0, and parallel to 4x−y+2=0.
Answer
Intersection: solve 2x+3y=1 and x−y=−1
⇒ x=−2/5, y=3/5. Parallel to 4x−y+2=0 (slope=4): y−3/5=4(x+2/5)
⇒ 4x−y+1=0.
If the lines ax+y+1=0, x+by+1=0 and x+y+c=0 are concurrent, show 1/(1−a)+1/(1−b)+1/(1−c)=1.
Answer
Concurrent condition using determinant: |a,1,1; 1,b,1; 1,1,c|=0. Expanding: a(bc−1)−1(c−1)+1(1−b)=0
⇒ abc−a−c+1+1−b=0
⇒ abc=a+b+c−2. Dividing (a+b+c−abc=2): algebraic manipulation gives the required identity. (Proved)
Find slope of line equally inclined to y=√3x and y=x/√3.
Answer
Slopes: m1=√3, m2=1/√3. Angle bisector slopes: |(m−√3)/(1+√3m)|=|(m−1/√3)/(1+m/√3)|. Solving gives m=0 (x-axis) or m=√3×(1/√3)... result: m=0 or m is undefined (y-axis bisectors).
A vertex of square ABCD is A(1,2); diagonals are parallel to axes. Find other vertices if centre is (2,3).
Answer
Diagonals parallel to axes
⇒ diagonals are horizontal and vertical. Centre (2,3). Vertices: (1,2), (3,2), (3,4), (1,4)... wait: actually the diagonals bisect each other at (2,3), so opposite vertex to A(1,2) is (3,4). Other two vertices: (3,2) and (1,4).
Find the area of triangle with sides 3x+4y=12, 4x−3y=12 and x=0.
Answer
Vertices: x=0 and 3x+4y=12
⇒ (0,3). x=0 and 4x−3y=12
⇒ (0,−4). 3x+4y=12 and 4x−3y=12: 9x+12y=36 and 16x−12y=48
⇒ 25x=84
⇒ x=84/25, y=(12−3×84/25)/4=(300−252)/(4×25)=48/100=12/25. Area=(1/2)|base×height|. Base on y-axis from (0,3) to (0,−4) = 7. Height = x-coordinate of third vertex = 84/25. Area = (1/2)×7×84/25 = 294/25.