Concurrent Lines - Test Questions

Question 1

Question:   Find the centroid of the triangle whose sides are
x + y - 1 = 0, x - 3y + 3 = 0 and x - y - 1 = 0.

Answer:    Let AB represent the side x + y - 1 = 0 ...(i)
Let BC represent the side x - 3y + 3 = 0 ... (ii)
Let CA represent the side x - y - 1 = 0 ... (iii)
Solving (i) and (ii) for x and y, we get the coordinates of B, i.e., B(0,1).
Solving (ii) and (iii) , we get the coordinates of C, i.e., C(3,2).
Solving (iii) and (i), we get the coordinates of A, i.e., A(1,0).
The coordinates of the centroid of the triangle are

Question 2

Question:   If the two vertices of a triangle are (3,-1) and (-2,3) and its orthocentre is the origin, find the coordinates of the third vertex.

Answer:   
In DABC, A(3, -1) and B(-2, 3) and the orthocentre is O(0, 0).
Let AD and BE be the altitudes through A and B.
The third vertex is the intersection of the lines BD and AE.






\ Slope of BD = 3



Solving (i) and (ii) for x and y, we get

Question 3

Question:   Find the equation of the line which joins the points (a,b) to the

Answer:    The two given equation are


The equation of the line through the intersection of (i) and (ii) is (bx + ay - ab) + l(ax + by - ab) = 0.
This line passes through (a,b).

Question 4

Question:   Prove that following lines are concurrent:
15x - 18y + 1 = 0, 12x + 10y - 3 = 0, 6x + 66y - 11 = 0.

Answer:    15x - 18y + 1 = 0 ...(i)
12x + 10y - 3 = 0 ...(ii)
6x + 66y - 11 = 0 ...(iii)
Solve (i) and (ii) for x and y,








0
RHS
Hence the given lines are concurrent.

Question 5

Question:  

Answer:    Let the equation of the straight line be y = mx + c or mx - y + c = 0.

Question 6

Question:   Show that for the line 5x + 6y + 12, (2,-3) and the origin are on the same side.

Answer:    Let d = distance between the origin and the line.

Let d' = distance between (2,-3) and the line.

Since d and d' are of the same sign, the origin and the point are on the same side of the line.

Question 7

Question:   Find the equation of the bisectors of the angles between the lines 3x - 4y = 8 and 4x + 3y = 10.

Answer:    The given equation of lines is rewritten with positive constant.
i.e., -3x + 4y + 8 = 0 and -4x - 3y + 10 = 0
The equation of the bisectors is


Taking only positive values, we get


Taking negative sign only, we get

Question 8

Question:   The distance of a point (x₁, y₁) from each of two straight lines, which pass through the origin is d. Prove that the two lines are given by (x₁y - xy₁)² = d (x² + y²).

Answer:    Let y = mx be any line through the origin. Then, distance d of (x₁, y₁)

Question 9

Question:   Find the equation of the line, which cuts off an intercept of 5 units on negative direction of y-axis and makes an angle of 135^o with the positive direction of x-axis.

Answer:    Here m = 135^o = tan (180^o - 45^o)
= - tan45^o
= -1
and c = -5
\ Required equation of the line is
y = (-1)x - 5
y = - x - 5 or x + y + 5 = 0

Question 10

Question:  

Answer:   


The equation of the line is of the form y = mx + c.

6y 2x - 3
6y - 2x + 3 0