Question

Find the value of 'a' if the distances of the points (2, 3) and (-4, a) from the straight line 3x + 4y - 8 = 0 are equal.

Answer 1

take care of the mod. any further doubt?? thn comment. ok!

Answer 2

Given:

The distances of the points (2, 3) and (-4, a) from the straight line 3x + 4y - 8 = 0 are equal

To find:

The value of a

Solution:

The required value of a is 7.5 or 2.5.

We are aware that the distance in any given point and a line is obtained using the following-

The required distance between a line and a point=| Ax1+By1+c| /$\sqrt{(A^{2} +B^{2}) }$

We have the line's equation as 3x + 4y - 8 = 0.

So, A=3, B=4, and C= -8.

We will calculate the two equal distances by substituting x1 and y1 with the points given.

Now, the distance in (2, 3) and ( 3x + 4y - 8=0)= | 3×2+4×3 -8|/$\sqrt{3^{2} +4^{2} }$

=| 6+12-8 |/ 5

=10/5

=2 units (1)

Similarly, the distance in (-4, a) and ( 3x + 4y - 8=0)= | 3×(-4)+4×a-8 |/$\sqrt{3^{2} +4^{2} }$

=| -12+4a-8|/5

=|4a-20|/5 units (2)

Since the two distance are equal, (1)=(2).

2=|4a-20|/5

10=|4a-20|

So, now (4a-20)=10 and (20-4a)=10

10=4a-20

30=4a

7.5=a

Also, 20-4a=10

10=4a

2.5=a

Thus, the value of a is 7.5 or 2.5.