Question

The slope of a straight line passing through A( -2, 3) is -4/3. The points on the line that are 10 units away from A are

• A. (-8, 11), (4, -5)
• B. (- 7, 9), (17,-1)
• C. (7, 5) (- 1, -1)
• D. (6, 10), (3, 5)

Answer 1

The correct option is A (-8, 11), (4, -5)

The equation line will be

$y-3=\frac{-4}{3}(x+2)$

$4x+3y=1$

Now, let the point $10$ unit away is $(h,k)$

Then,

$4h+3k=1$ ------ $\left(1\right)$

And

$\sqrt{(h+2{)}^2+(k-3{)}^2}=10$

$(h+2{)}^2+(k-3{)}^2=10$ ------ $\left(2\right)$

Now, from $\left(1\right)$

$h=\frac{1-3k}{4}$

Put in equation $\left(2\right)$, we get

$25k^2-150k-1663=0$

On solving quadric equation

$K=11$ and $k=-5$

Put $k=11$ in equation $\left(1\right)$

$h=-8$

And put $k=-5$ in equation $\left(1\right)$

$h=4$

Hence the points are

$(-8,11),(4,-5)$

Hence, the correct answer is $A$.