Question

question is in photo​

Answer 1

Answer:

Point Q is$(\frac{-1}{2},0 )$.

The given triangle is an isosceles triangle.

Step-by-step explanation:

(a) Let Q(x,0) be the unknown point.

Given: that Q(x,0) is equidistant from A(-5,-2) and B(4,-2)

So AQ = QB

=> $AQ^{2} = QB^{2}$

Therefore, By distance formula

$(x+5)^{2} + (0+2)^{2} = (x-4)^{2} + (0+2)^{2}\\x^{2}+10x+25= x^{2}-8x+16\\10x+8x=16-25\\18x = -9\\x = \frac{-1}{2}$

Hence, the coordinates of point Q is$(\frac{-1}{2},0 )$.

(b) Let QAB be a triangle with coordinates Q $(\frac{-1}{2},0 )$ , A(-5,-2) and B(4,-2)  

Given that AQ=BQ

Therefore the given triangle is an isosceles triangle.

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