Question

Equation of straight line whose slope is 3 and which bisects the joint of (-2,5) and (3,4) is?"

Answer 1

The required line bisects the joint of (-2,5) and (3,4)
So the point of contact is given by (h,k)

h = (- 2 + 3)/2 = 1/2
k = (5 + 4)/2 = 9/2

So the common point is (1/2,9/2)

Let the required equation be y = mx + c

Here, slope m = 3
y = 3x + c

Since (1/2,9/2) is a solution of y = 3x + c ,so

9/2 = 3/2 + c
c = 6/2 = 3

Now, the required straight line equation is
y = 3x + 3
or 3x - y + 3 = 0

Hope This Helps You!

Answer 2

We have to find the line which bisects the joint of given points (-2,5) and (3,4).

So, The x coordinate of point at which bisection happens is (-2+3)/2=1/2

The y co-ordinate is (5+4)/2=9/2



We know that the point of bisection or contact is (1/2,9/2) which must happen to pass through this line.

Hence, (1/2,9/2) is a solution of the required line.

Now, We know general form of a line with slope m is y = mx + c

Now, Given slope = 3.

y =3x + c

9/2=3(1/2)+c

9/2-3/2=c
6/2=3=c

Now, The equation of line is y =3x+3

3x -y + 3=0 is the required line

Hope this helps :)