Question

find the point on y-axis each of which is at a distance of 13 units from the point (-5,7)

Answer 1

Step-by-step explanation:

√(x+5)²+(y-7)² = 13. [Distance Formula]

x²+25+10x + y²+49–14y=169 [After Squaring both sides]

x²+y²+10x-14y-95=0

For y axis put x=0 in above eqn:

We get: y²-14y-95=0

y²-19y+5y-95=0

y(y-19) +5(y-19)=0

(y-19)(y+5)=0

y=19 , -5

So 19 and -5 are the required points..