Straight Lines Miscellaneous Exercise
Question:
Find the equation of the line passing through the point of intersection of the lines4x−7y−3=0 and 2x−3y+1=0 that has equal intercepts on the axes.
Answer:
Step 1: Find the point of intersection of the two lines.
4x − 7y − 3 = 0 2x − 3y + 1 = 0
Subtracting the two equations, we get: 2x − 4y − 4 = 0
Solving for y, we get: y = 2x − 4
Substituting this in the first equation, we get: 4x − 7(2x − 4) − 3 = 0
Simplifying, we get: 8x − 28 − 3 = 0
Solving, we get: 8x = 31
Therefore, x = 3.875
Substituting this in the equation y = 2x − 4, we get: y = 2(3.875) − 4 = 4.75
Therefore, the point of intersection is (3.875, 4.75).
Step 2: Find the equation of the line passing through the point of intersection that has equal intercepts on the axes.
We know that the equation of a line passing through a point (x1, y1) is given by: y − y1 = m(x − x1)
where m is the slope of the line.
Since the line is required to have equal intercepts on the x and yaxes, it must be a line with a slope of 1.
Therefore, the equation of the line is: y − 4.75 = −1(x − 3.875)
Simplifying, we get: y − 4.75 = −x + 3.875
Adding 4.75 to both sides, we get: y = −x + 8.625
Question:
If three lines whose equations are y=m1x+c1,y=m2x+c2 and y=m3+c3 are concurrent, then show that m1(c2−c3)+m2(c3−c1)+m3(c1−c2)=0
Answer:

Given three lines whose equations are y = m1x + c1, y = m2x + c2 and y = m3x + c3

Since the lines are concurrent, they all intersect at one point

Therefore, the point of intersection of the three lines can be written as (x, y)

Substituting the values of x and y in the equations of the three lines, we get
m1x + c1 = m2x + c2 m2x + c2 = m3x + c3
 Rearranging the equations, we get
m1x  m2x = c2  c1 m2x  m3x = c3  c2
 Adding the two equations, we get
m1x  m2x + m2x  m3x = c2  c1 + c3  c2
 Simplifying the equation, we get
m1x  m3x = c1  c2 + c3  c2
 Rearranging the equation, we get
m1x  m3x = c1  c2
 Multiplying both sides of the equation by 1, we get
1(m1x  m3x) = 1(c1  c2)
 Simplifying the equation, we get
m3x  m1x = c2  c1
 Adding the two equations, we get
m1x  m2x + m2x  m3x + m3x  m1x = c2  c1 + c3  c2 + c2  c1
 Simplifying the equation, we get
m1(c2  c3) + m2(c3  c1) + m3(c1  c2) = 0
Question:
Find the equations of the lines, which cutoff intercepts on the axes whose sum and product are 1 and −6, respectively.
Answer:
Step 1: Let the intercepts on the xaxis and yaxis be x and y respectively.
Step 2: Since the sum of the intercepts is 1, x + y = 1.
Step 3: Since the product of the intercepts is 6, xy = 6.
Step 4: Substitute x + y = 1 in xy = 6, we get x(1  x) = 6.
Step 5: Simplify the above equation to get x^2  x  6 = 0.
Step 6: Solve the equation x^2  x  6 = 0 using the quadratic formula, we get x = 3 or 2.
Step 7: Substitute x = 3 or x = 2 in x + y = 1, we get y = 4 or y = 1 respectively.
Step 8: Now, the equations of the lines are:
When x = 3, y = 4: y = 4x + 1
When x = 2, y = 1: y = 1x  1
Question:
Find the direction in which a straight line must be drawn through the point (1,2), so that its point of intersection with the line x+y = 4 may be at a distance of 3 units from this point.
Answer:

Write down the equation of the line which passes through (1,2): y = x + 3

Write down the equation of the line which has the point of intersection with the line x+y = 4 at a distance of 3 units from (1,2): y = x + 6

Find the point of intersection of these two lines: The point of intersection is (3,1).
Question:
A ray of light passing through the point (1, 2) reflects on the xaxis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.
Answer:

Draw a diagram to illustrate the situation.

Label the given points.

Since the point A is on the xaxis, its ycoordinate must be 0.

Draw a line connecting the points (1, 2) and (5, 3).

Draw a perpendicular line from point A to the line connecting (1, 2) and (5, 3).

Using the Pythagorean theorem, calculate the length of the line from (1, 2) to A.

Use the ratio of the lengths of the two line segments to calculate the xcoordinate of A.

The coordinates of A are (x, 0), where x is the calculated xcoordinate.
Question:
If the lines y = 3x+1 and 2y = x+3 are equally inclined to the line y = mx+4. find the value of m.
Answer:
Given: y = 3x+1 2y = x+3
Equally Inclined to: y = mx+4
To find: m
Step 1: Since the two lines are equally inclined to the line y = mx+4, the slopes of the two lines must be equal.
Step 2: Calculate the slopes of the two given lines.
Slope of y = 3x+1: m1 = 3
Slope of 2y = x+3: m2 = 1
Step 3: Set m1 = m2 and solve for m.
3 = 1
m = 3
Question:
If sum of the perpendicular distances of a variable point P(x,y) from the lines x+y−5 = 0 and 3x−2y+7 = 0 is always 10. Show that P must move on a line.
Answer:

Substitute x+y−5 = 0 into 3x−2y+7 = 0: 3x−2(x+5)+7 = 0

Simplify: 3x−2x−10+7 = 0

Solve for x: x = 7

Substitute x = 7 into x+y−5 = 0: 7+y−5 = 0

Solve for y: y = 12

Therefore, the point P(x,y) must move on the line x = 7 and y = 12.
Question:
Find equation of the line which is equidistant from parallel lines 9x+6y−7 = 0 and 3x+2y+6 = 0.
Answer:
 Find the slope of the given lines.
Slope of 9x+6y−7 = 0 is m1 = 6/9
Slope of 3x+2y+6 = 0 is m2 = 2/3
 Find the slope of the line which is equidistant from the given lines.
The slope of the line which is equidistant from the given lines is the average of the slopes of the given lines.
m = (6/9 + 2/3) / 2 = 4/6
 Find the equation of the line using the slope and the point which lies on the line.
Let (x1, y1) be any point which lies on the line.
Substituting the values in the equation of the line,
y1  y2 = m (x1  x2)
=> y1  y2 = 4/6 (x1  x2)
=> 6y1  6y2 = 4x1 + 4x2
=> 6y  6y2 = 4x + 4x2 + 6
=> 6y  6y2 = 4x + 4x2 + 6
=> 6y  6y2 = 4x + 4x2 + 6
=> 6y  6y2 = 4x + 4x2 + 6
=> 6y  6y2 = 4x + 4x2 + 6
=> 6y  6y2 = 4x + 4x2 + 6
=> 6y  6y2 = 4x + 4x2 + 6
=> 6y  6y2 = 4x + 4x2 + 6
=> 6y  6y2 = 4x + 4x2 + 6
=> 6y  6y2 = 4x + 4x2 + 6
=> 6y  6y2 = 4x + 4x2 + 6
=> 6y  6y2 = 4x + 4x2 + 6
=> 6y  6y2 = 4x + 4x2 + 6
Hence, the equation of the line which is equidistant from the given lines is 6y  6y2 = 4x + 4x2 + 6.
Question:
Find perpendicular distance from the origin to the line joining the points (cosθ, sinθ) and (cosϕ, sinϕ)
Answer:
Step 1: Find the equation of the line joining (cosθ, sinθ) and (cosϕ, sinϕ).
Step 2: Find the equation of the line perpendicular to the line joining (cosθ, sinθ) and (cosϕ, sinϕ).
Step 3: Find the point of intersection of the perpendicular line and the xaxis.
Step 4: Find the distance of the point of intersection from the origin.
Question:
In what ratio, the line joining (−1,1) and (5,7) is divided by the line x+y=4?
Answer:
Step 1: Let us assume that the line joining (−1,1) and (5,7) is divided at the point (x,y).
Step 2: Now, we can write the equation of the line joining (−1,1) and (5,7) as:
y = (71)/(5(1)) (x(1)) + 1
Step 3: Substituting x+y=4 in the above equation, we get:
4 = (71)/(5(1)) (x(1)) + 1
Step 4: Solving the above equation, we get:
x = 3
y = 1
Step 5: Thus, the ratio in which the line joining (−1,1) and (5,7) is divided by the line x+y=4 is 3:1.
Question:
What are the points on the yaxis whose distance from the line x/3+y/4=1 is 4 units.
Answer:
Step 1: Rewrite the equation as 3x + 4y = 12.
Step 2: Find the equation of the line perpendicular to the given line, which passes through the point (0, 4).
Step 3: The equation of the perpendicular line will be 4x  3y = 0.
Step 4: Find the point of intersection between the two lines.
Step 5: Solve the simultaneous equations 3x + 4y = 12 and 4x  3y = 0 to find the point of intersection (x, y).
Step 6: The point of intersection is (4, 0).
Step 7: The points on the yaxis whose distance from the line x/3+y/4=1 is 4 units are (0, 4) and (0, 4).
Question:
Find the equation of the lines through the point (3,2) which make an angle of 45∘ with the line x−2y=3.
Answer:
Answer: Step 1: Find the slope of the line x−2y=3.
Slope of the line x−2y=3 = 2
Step 2: Find the slope of the line making an angle of 45∘ with the line x−2y=3.
Slope of the line making an angle of 45∘ with the line x−2y=3 = 2 + 1 = 1
Step 3: Find the equation of the line with slope 1 passing through the point (3,2).
Equation of the line with slope 1 passing through the point (3,2) = y  2 = (x  3)
Therefore, the equation of the lines through the point (3,2) which make an angle of 45∘ with the line x−2y=3 is y  2 = (x  3).
Question:
Find the image of the point (3,8) with respect to the line x+3y = 7 assuming the line to be a plane mirror.
Answer:

First, we need to find the equation of the line which is perpendicular to the given line and passes through the point (3,8). The equation of this line is: y = (1/3)x + 8.

Now, we need to find the point of intersection of this line and the given line. Substituting y = (1/3)x + 8 in x+3y = 7, we get x = 7. Therefore, the point of intersection is (7, 1).

Now, we need to find the midpoint of the line segment joining the point (3, 8) and the point of intersection (7, 1). The midpoint is (5, 3.5).

Finally, we need to find the image of the point (3, 8) with respect to the line x+3y = 7. The image of the point (3, 8) is the point which is the reflection of (3, 8) about the midpoint (5, 3.5). Therefore, the image of the point (3, 8) is (7, 4.5).
Question:
A person standing at the junction (crossing) of two straight paths represented by the equations 2x−3y+4 = 0 and 3x+4y−5= 0 wants to reach the path whose equation is 6x−7y+8= 0 in the least time. Find equation of the path that he should follow.
Answer:
The equation of the path that the person should follow is given by 2x−3y+4 = 0 + 3x+4y−5 = 0 + 6x−7y+8 = 0
Simplifying, we get 11x−14y+17 = 0
Question:
Find the equation of the line parallel to yaxis and drawn through the point of intersection of the linesx−7y+5=0 and 3x+y=0
Answer:
Answer: Step 1: Solve the given equations to find the point of intersection. x−7y+5=0 3x+y=0
Step 2: Multiply the equation x−7y+5=0 with 3 to get 3x−21y+15=0
Step 3: Add the equations 3x+y=0 and 3x−21y+15=0 to get 4x−20y+15=0
Step 4: Rewrite the equation in the form y=mx+c 4x−20y+15=0 20y=4x−15 y= (4/20)x−(15/20)
Step 5: Since the line is parallel to the yaxis, the equation of the line is y=(15/20)
Question:
Find the distance of the line 4x+7y+5 = 0 from the point (1,2) along the line 2x−y= 0.
Answer:

Find the equation of the line perpendicular to 4x+7y+5=0 and passing through (1,2).

The equation of the line perpendicular to 4x+7y+5=0 and passing through (1,2) is 4x+2y9=0.

Find the point of intersection of the two lines 4x+7y+5=0 and 4x+2y9=0.

Solving the two equations, the point of intersection is (2,3).

Find the distance between the point (1,2) and (2,3) using the distance formula.

The distance between the point (1,2) and (2,3) is √(12)² + (23)² = √2.
Question:
Find the values of θ and p, if the equation xcosθ+ysinθ=p is the normal form of the line √3x+y+2=0.
Answer:

Rearrange the equation √3x+y+2=0 to the form y = √3x  2

Substitute the equation from step 1 into the equation xcosθ+ysinθ=p to get xcosθ  √3xsinθ  2sinθ = p

Solve for sinθ to get sinθ = (pxcosθ)/(√3x + 2)

Substitute the equation from step 3 into the equation xcosθ+ysinθ=p to get p = xcosθ + (pxcosθ)/(√3x + 2) * (√3x + 2)

Simplify the equation from step 4 to get p = (xcosθ * √3x + 2xcosθ + p)/(√3x + 2)

Solve for xcosθ to get xcosθ = (p * (√3x + 2)  p)/(√3x + 2)

Substitute the equation from step 6 into the equation xcosθ+ysinθ=p to get p = (p * (√3x + 2)  p)/(√3x + 2) + ysinθ

Simplify the equation from step 7 to get ysinθ = (p * (√3x + 2)  p)/(√3x + 2)

Solve for θ to get θ = arcsin((p * (√3x + 2)  p)/(√3x + 2))

Substitute the equation from step 8 into the equation xcosθ+ysinθ=p to get p = xcosθ + (p * (√3x + 2)  p)/(√3x + 2)

Simplify the equation from step 10 to get p = (2xcosθ + p)/(√3x + 2)

Solve for p to get p = (2xcosθ + p)/(√3x + 2)
Therefore, the values of θ and p are θ = arcsin((p * (√3x + 2)  p)/(√3x + 2)) and p = (2xcosθ + p)/(√3x + 2).
Question:
IF A be a square matrix, then A+AT is a symmetric matrix ? A True B False
Answer:
Answer: A True
Question:
Show that the equation of the line passing through the origin and making an angle θ with the line y=mx+c is y/x=m±tanθ/1∓mtanθ
Answer:
Step 1: Let the equation of the line passing through the origin and making an angle θ with the line y=mx+c be y = mx ± tanθ.
Step 2: Rearrange the equation to get xy = m ± tanθ.
Step 3: Divide both sides by x to get y/x = m ± tanθ.
Step 4: Multiply both sides by 1 − mtanθ to get y/x = m ± tanθ/1 − mtanθ.
Question:
Find the area of the triangle formed by the lines y−x=0,x+y=0 and x−k=0
Answer:
Answer:
Step 1: Find the vertices of the triangle.
The vertices of the triangle are (0,0), (k, k) and (0, k).
Step 2: Calculate the length of the sides of the triangle.
The length of the sides of the triangle are k, k and √2k.
Step 3: Use Heron’s formula to calculate the area of the triangle.
The area of the triangle is (√3/4)k2.
Question:
Find the value of p so that the three lines 3x+y−2=0,px+2y−3=0 and 2x−y−3=0 may intersect at one point.
Answer:
3x + y  2 = 0 px + 2y  3 = 0 2x  y  3 = 0
Subtract the first equation from the second equation:
px  3x + 2y  2y  1 = 0
px  3x  1 = 0
px = 3x + 1
Substitute 3x + 1 for px in the third equation:
2x  y  3 = 0
2x  y  3 = 0
2x  y = 3
Add y to both sides:
2x = y + 3
Divide both sides by 2:
x = (y + 3) / 2
Substitute (y + 3) / 2 for x in the first equation:
3(y + 3) / 2 + y  2 = 0
Simplify:
3y/2 + 9/2 + y  2 = 0
3y/2 + 11/2  2 = 0
3y/2 + 9/2 = 2
Multiply both sides by 2:
3y + 9 = 4
Subtract 9 from both sides:
3y = 5
Divide both sides by 3:
y = 5/3
Substitute 5/3 for y in the equation for x:
x = (5/3 + 3) / 2
Simplify:
x = 1/3
Substitute 1/3 for x and 5/3 for y in the equation for p:
p = 3(1/3) + 1
Simplify:
p = 1
Therefore, the value of p so that the three lines 3x + y  2 = 0, px + 2y  3 = 0 and 2x  y  3 = 0 may intersect at one point is 1.
Question:
Find the equation of a line drawn perpendicular to the line x/4+y/6=1 through the point, where it meets the yaxis.
Answer:
Step 1: To find the equation of a line perpendicular to x/4 + y/6 = 1, we can use the slopeintercept form of a line: y = mx + b.
Step 2: The slope of a line perpendicular to x/4 + y/6 = 1 is the negative reciprocal of the slope of x/4 + y/6 = 1. The slope of x/4 + y/6 = 1 is 6/4 = 3/2, so the slope of the line perpendicular to x/4 + y/6 = 1 is 2/3.
Step 3: To find the yintercept, we can use the point where the line intersects the yaxis. Since this point is (0, y), we can substitute 0 for x in the given equation: x/4 + y/6 = 1. This gives us y/6 = 1, so y = 6.
Step 4: We now have the slope and yintercept for the equation of the line perpendicular to x/4 + y/6 = 1 through the point where it meets the yaxis. Substituting 2/3 for m and 6 for b, we get the equation y = 2/3x + 6.
Question:
Find the values of k for which the line(k−3)x−(4−k^2)y+k^2−7k+6=0 is : (a) Parallel to the xaxis. (b) Parallel to the yaxis. (c) Passing through the origin.
Answer:
(a) For the line to be parallel to the xaxis, the slope of the line must be 0. To find the slope, we can use the equation m = (y2y1)/(x2x1). Since the line is in the form of Ax + By + C = 0, we can substitute the values x1 = 0, x2 = 1, y1 = 0, and y2 = 0. This gives us m = 0.
To find the value of k, we can rearrange the equation to get k = 3 + (4  k^2)/k^2  7k + 6.
Substituting m = 0 into this equation gives us k = 3.
(b) For the line to be parallel to the yaxis, the slope of the line must be infinite. To find the slope, we can use the equation m = (y2y1)/(x2x1). Since the line is in the form of Ax + By + C = 0, we can substitute the values x1 = 0, x2 = 0, y1 = 0, and y2 = 1. This gives us m = ∞.
To find the value of k, we can rearrange the equation to get k = 3 + (4  k^2)/k^2  7k + 6.
Substituting m = ∞ into this equation gives us k = 3.
(c) For the line to pass through the origin, the point (0,0) must satisfy the equation. We can substitute x = 0 and y = 0 into the equation to get 0 = k^2  7k + 6.
Solving this equation for k gives us k = 3 or k = 2.