Straight Lines Exercise 03
Question:
If 2p is the length of perpendicular from the origin to the line whose intercepts on the axes are 2a and b, then show that 1/p^2=1/a^2+1/b^2.
Answer:

Given: 2p = length of perpendicular from the origin to the line whose intercepts on the axes are 2a and b

We need to show that: 1/p2 = 1/a2 + 1/b2

We can rewrite the equation as: p2 = a2 + b2

Taking the reciprocal of both sides of the equation: 1/p2 = 1/a2 + 1/b2

Therefore, 1/p2 = 1/a2 + 1/b2, which is the desired result.
Question:
The perpendicular from the origin to the line y =mx+c meets it at the point (−1, 2). Find the values of m and c.
Answer:

First, we need to solve for m. We can do this by substituting the point (1, 2) into the equation y = mx + c and solving for m.

1 = m(1) + c

2 = m(0) + c

Subtracting the second equation from the first equation gives us:

1 = m + c

Adding m to both sides gives us:

m = 1 + c

Substituting this equation into the second equation gives us:

2 = (1 + c)(0) + c

2 = c

Therefore, m = 1 + c and c = 2.
title: “03” parent: “10 Straight Lines Exercise” draft: false
Question:
Reduce the following equations into slope  intercept form and find their slopes and the y  intercepts. (i)x+7y =0 (ii)6x+3y−5 =0 (iii)y =0
Answer:
(i) x + 7y = 0 Slope: 7 Yintercept: 0
(ii) 6x + 3y  5 = 0 Slope: 3/6 Yintercept: 5/3
(iii) y = 0 Slope: 0 Yintercept: 0
Question:
If p and q are the lengths of perpendiculars from the origin to the lines xcosθ−ysinθ=kcos2θ and xsecθ+ycscθ=k, respectively, prove that p^2+4q^2=k^2
Answer:
 xcosθ−ysinθ=kcos2θ
 xsecθ+ycscθ=k
 Take the square of the first equation: (xcosθ−ysinθ)^2=k^2cos^22θ
 Take the square of the second equation: (xsecθ+ycscθ)^2=k^2
 Multiply the first equation by the second equation: (xcosθ−ysinθ)(xsecθ+ycscθ) = k^2cos^22θ
 Expand the product on the left side of the equation: x^2cosθsecθ−xycosθcscθ−yxsecθsinθ+y^2sinθcscθ = k^2cos^22θ
 Simplify the left side of the equation: x^2cosθsecθ−xy(cosθcscθ+secθsinθ)+y^2sinθcscθ = k^2cos^22θ
 Factor out x^2 and y^2 from the left side of the equation: x^2(cosθsecθ−(cosθcscθ+secθsinθ))+y^2(sinθcscθ) = k^2cos^22θ
 Simplify the factors on the left side of the equation: x^2(cosθ(secθ−cscθ−sinθsecθ))+y^2(sinθ(cscθ−sinθcscθ)) = k^2cos^22θ
 Simplify the factors on the left side of the equation further: x^2(−sinθ)+y^2(−cosθ) = k^2cos^22θ
 Take the square root of both sides of the equation: √(x^2(−sinθ)+y^2(−cosθ)) = √(k^2cos^22θ)
 Simplify the square root on the left side of the equation: √(x^2)+√(y^2) = √(k^2cos^22θ)
 Substitute p and q for the square roots on the left side of the equation: p+q = √(k^2cos^22θ)
 Square both sides of the equation: (p+q)^2 = k^2cos^22θ
 Expand the square on the left side of the equation: p^2+2pq+q^2 = k^2cos^22θ
 Simplify the left side of the equation: p^2+2pq+q^2 = k^2cos^22θ
 Subtract 2pq from both sides of the equation: p^2+q^2 = k^2cos^22θ−2pq
 Factor out q^2 from the left side of the equation: q^2(1+p^2/q^2) = k^2cos^22θ−2pq
 Simplify the equation: q^2(1+p^2/q^2) = k^2cos^22θ−2pq
 Divide both sides of the equation by (1+p^2/q^2): q^2 = (k^2cos^22θ−2pq)/(1+p^2/q^2)
 Square both sides of the equation: q^4 = (k^2cos^22θ−2pq)^2/(1+p^2/q^2)^2
 Simplify the equation: q^4 = (k^4cos^44θ−4k^2pqcos^22θ+4p^2q^2)/(1+p^2/q^2)^2
 Add 4p^2q^2 to both sides of the equation:
Question:
Find the perpendicular distance of the point (−1,1) from the line 12(x+6)=5(y−2).
Answer:
Step 1: Find the equation of the line in the form of ax + by = c.
12(x+6)=5(y−2) 12x + 72 = 5y  10 5y = 12x + 82 y = (12/5)x + (82/5)
Step 2: Find the equation of the perpendicular line.
Slope of the perpendicular line = (1/m) m = (12/5) So, slope of perpendicular line = (5/12)
Step 3: Find the equation of the perpendicular line using the point (−1,1).
Let (x1,y1) = (−1,1) Slope of perpendicular line = (5/12) Equation of the perpendicular line is y  y1 = (5/12)(x  x1) y  1 = (5/12)(x + 1) y = (5/12)x  (7/12)
Step 4: Find the intersection of the two lines.
Equation of line: y = (12/5)x + (82/5) Equation of perpendicular line: y = (5/12)x  (7/12)
Solving the two equations simultaneously: (12/5)x + (82/5) = (5/12)x  (7/12) (17/12)x = (89/12) x = (89/17)
Substitute x = (89/17) in the equation of the line: y = (12/5)(89/17) + (82/5) y = (1089/85)
Step 5: Find the distance between the two points.
Let (x1,y1) = (−1,1) Let (x2,y2) = (89/17, 1089/85) Distance = √((x2−x1)2+(y2−y1)2) Distance = √(((89/17)−(−1))2+((1089/85)−(1))2) Distance = √((90/17)2+((1088/85)2) Distance = √((90×17/17×17)+(1088×85/85×85)) Distance = √((90/289)+(1088/7225)) Distance = √(90×7225/289×7225 + 1088/7225) Distance = √(64500/2077225 + 1088/7225) Distance = √(64500/2077225 + 1088/7225) Distance = √(7.133) Distance = 2.67
Question:
Find the equation of the line parallel to the line 3x−4y+2=0 and passing through the point (−2,3).
Answer:
Answer: Step 1: Identify the slope of the given line, 3x−4y+2=0
Step 2: Calculate the slope of the given line, m = 3/4
Step 3: Identify the coordinates of the given point, (2,3)
Step 4: Calculate the slope of the line parallel to the given line, m = 3/4
Step 5: Use the pointslope form of the equation of a line, y  3 = 3/4(x + 2)
Step 6: Simplify the equation, 4y  12 = 3x  6
Step 7: Rearrange the equation to standard form, 3x + 4y  12 = 0
Therefore, the equation of the line parallel to the line 3x−4y+2=0 and passing through the point (−2,3) is 3x + 4y  12 = 0.
Question:
Find the equation of the right bisector of the line segment joining the points (3,4) and (−1,2).
Answer:
Step 1: Find the midpoint of the line segment joining the points (3,4) and (−1,2).
The midpoint can be found by taking the average of the xcoordinates and the average of the ycoordinates.
Midpoint = ( (3 + (1)) / 2 , (4 + 2) / 2 )
Midpoint = (1, 3)
Step 2: Find the slope of the right bisector.
The slope of the right bisector is the negative reciprocal of the slope of the line joining the two points.
Slope of line joining the two points = (2  4) / (1  3)
Slope of line joining the two points = 2
Slope of right bisector = 1/2
Step 3: Find the equation of the right bisector.
The equation of the right bisector can be found using the slope and the midpoint.
Equation of right bisector: y  3 = 1/2 (x  1)
Simplifying: y = 1/2 x + 5/2
Question:
If p is the length of perpendicular from the origin to the line whose intercepts on the axis are a and b, then show that 1/p^2=1/a^2+/b^2.
Answer:
Step 1: Recall that the equation of a line passing through the origin with intercepts a and b is y = ax + b.
Step 2: Let P be the length of the perpendicular from the origin to the line y = ax + b.
Step 3: Draw a right triangle with the origin as the vertex, the line y = ax + b as the hypotenuse, and P as the perpendicular side.
Step 4: Using the Pythagorean Theorem, we can write P^2 = a^2 + b^2.
Step 5: Rearrange the equation to get 1/P^2 = 1/a^2 + 1/b^2.
Question:
Two lines passing through the point (2,3) intersects each other at an angle of 60∘. If slope of one line is 2, find equation of the other line.
Answer:

Slope of the other line = 1/2

Equation of the line is y = 1/2x + c

Substitute (2,3) in the equation to get c = 7

Final equation of the line is y = 1/2x + 7
Question:
Reduce the following equations into intercept form and find their intercepts on the axes. (i)3x+2y−12=0 (ii)4x−3y=6 (iii)3y+2=0
Answer:
(i) 3x + 2y  12 = 0 => 2y = 3x  12 => y = 3/2x  6 Intercepts: (0, 6) and (4, 0)
(ii) 4x  3y = 6 => 3y = 4x  6 => y = 4/3x  2 Intercepts: (0, 2) and (3/2, 0)
(iii) 3y + 2 = 0 => 3y = 2 => y = 2/3 Intercepts: (0, 2/3) and (2/3, 0)
Question:
Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive xaxis. (i)x√(3)y+8=0 (ii)y−2=0 (iii)x−y=4
Answer:
(i) Standard form: y = (1/√(3))x  (8/√(3))
Perpendicular distance from the origin: 8/√(3)
Angle between perpendicular and the positive xaxis: tan1(1/√(3))
(ii) Standard form: y = 2
Perpendicular distance from the origin: 2
Angle between perpendicular and the positive xaxis: 90°
(iii) Standard form: y = x  4
Perpendicular distance from the origin: 4
Angle between perpendicular and the positive xaxis: 45°
Question:
Find the points on the xaxis, whose distances from the line x/3+y/4=1 are 4 units.
Answer:
Step 1: Find the equation of the line x/3 + y/4 = 1.
Step 2: Find the equation of the perpendicular line to the given line.
Step 3: Find the equation of the circle with centre on the perpendicular line and radius 4 units.
Step 4: Find the points of intersection of the circle and the xaxis.
Step 5: These points are the required points.
Question:
Find the angles between the lines √3 x + y = 1 and x + √3 y = 1.
Answer:
Step 1: Rewrite the equations in the form of y = mx + c, where m is the slope of the line.
Equation 1: y = √3x + 1 Equation 2: y = x/√3 + 1
Step 2: Find the slopes of the two lines.
Slope of equation 1 = √3 Slope of equation 2 = 1/√3
Step 3: Find the angle between the two lines using the formula: θ = arctan (m1  m2/1 + m1m2)
θ = arctan(√3  1/√3/1  √3/√3)
Step 4: Substitute the values in the formula and calculate the angle.
θ = arctan(√3 + 1/√3/1 + √3/√3)
θ = arctan(2/2√3)
θ = arctan(1/√3)
θ = 60°
Question:
The line through the points (h,3) and (4,1) intersects the line 7x−9y−19=0 at right angle. Find the value of h.
Answer:
 Write the equation of the line through the points (h,3) and (4,1) in slopeintercept form.
y = (1/3)x + (10/3)
 Find the slope of the line 7x−9y−19=0.
m = 9/7
 Find the slope of the line perpendicular to the line 7x−9y−19=0.
m = 7/9
 Substitute the slope of the perpendicular line into the equation of the line through the points (h,3) and (4,1).
(7/9)x + (10/3) = y
 Solve for h.
h = 9/2
Question:
Prove that the line through the point (x1,y1) and parallel to the line Ax+By+C=0. A(x−x1)+B(y−y1)=0
Answer:

Let (x1,y1) be a point on the line Ax+By+C=0.

Then, A(x−x1)+B(y−y1)=0.

Now, let (x2,y2) be a point on the line through (x1,y1) and parallel to the line Ax+By+C=0.

Then, A(x2−x1)+B(y2−y1)=0.

Subtracting A(x−x1)+B(y−y1)=0 from A(x2−x1)+B(y2−y1)=0, we get A(x2−x)+B(y2−y)=0.

Hence, the line through the point (x1,y1) and parallel to the line Ax+By+C=0 is A(x−x1)+B(y−y1)=0.
Question:
Find the distance between the parallel lines 15x + 8 y  34 = 0, 15 x +8y +31 = 0.
Answer:
Answer: Step 1: Subtract the equations to get: 15x + 8y  34 = 0 15x  8y  31 = 0
Step 2: Simplify to get: 3 = 0
Step 3: Since 3 is not equal to 0, the two parallel lines do not intersect and thus have no distance between them.
Question:
Find the equation of the line perpendicular to x−7y+5=0 and having x−intercept 3.
Answer:
Step 1: We know that the equation of a line perpendicular to x−7y+5=0 is 7x+y−5=0.
Step 2: To find the xintercept, let y=0 in the equation 7x+y−5=0.
Step 3: This gives us 7x−5=0.
Step 4: Solving for x gives us x=\frac{5}{7}.
Step 5: Since the xintercept is 3, we can set x=3 in the equation 7x+y−5=0.
Step 6: This gives us 21+y−5=0.
Step 7: Solving for y gives us y=16.
Step 8: Therefore, the equation of the line perpendicular to x−7y+5=0 and having x−intercept 3 is 7x+16−5=0, or 7x+11=0.
Question:
In the triangle ABC with vertices A(2,3),B(4,−1) and C(1,2), find the equation and length of altitude from the vertex A.
Answer:
Equation of altitude from vertex A: y3= (21)/(3(1))(x2) y3= (1)/(4)(x2) y3= (1/4)x (2/4) 4y12= x2 4yx= 14
Length of altitude from vertex A: Using distance formula: d = √((x2x1)^2 + (y2y1)^2) d = √((21)^2 + (3(1))^2) d = √(1^2 + 4^2) d = √17 d = 4.12