Question

Find a point P on the Y-axis which is equidistant from A(4,8) and B (-6,6). Also, find the distance of AP
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Answer 1

Answer:

(-1, 7)

√26

Step-by-step explanation:

As the point P is equidistant from A(4,8) and B (-6,6), means P is the mid point of the line join A and B. Using mid point formula,

=>Co-ordinates of P = ( (4+(-6))/2 ,(8+6)/2 )

=> Co-ordinates of P = ( -2/2 , 14/2 )

=> Co-ordinates of P = ( - 1, 7 )

Using distance formula :

AP = √(4 - (-1) )² + (8 - 7)²

AP = √5² + 1²

AP = √26

Answer 2

Given :-

• Points is A(4 , 8) and B(- 6 , 6).

To Find :-

• What is a point P on the Y-axis and also find the distance of AP.

Formula Used :-

$\clubsuit$ Mid-point Formula :

$\longmapsto \sf\boxed{\bold{\pink{Mid-point =\: \bigg(\dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2}\bigg)}}}$

$\clubsuit$ Distance Formula :

$\longmapsto \sf\boxed{\bold{\pink{Distance =\: \sqrt{{(x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2}}}}}$

Solution :-

First, we have to find the the point P :

Given :

• x₁ = 4
• x₂ = - 6
• y₁ = 8
• y₂ = 6

According to the question by using the formula we get,

$\implies \sf P =\: \bigg(\dfrac{4 + (- 6)}{2} , \dfrac{8 + 6}{2}\bigg)$

$\implies \sf P =\: \bigg(\dfrac{4 - 6}{2} , \dfrac{14}{2}\bigg)$

$\implies \sf P =\: \bigg(\dfrac{- \cancel{2}}{\cancel{2}} , \dfrac{\cancel{14}}{\cancel{2}}\bigg)$

$\implies \sf\bold{\red{P =\: (- 1 , 7)}}$

$\therefore$ The point P is (- 1 , 7) .

$\rule{150}{2}$

Now, we have to find the distance between AP :

Given :

$\mapsto$ A(4 , 8)

$\mapsto$ P(- 1 , 7)

Then,

• x₁ = - 1
• x₂ = 4
• y₁ = 7
• y₂ = 8

According to the question by using the formula we get,

$\implies \sf AP =\: \sqrt{{(4 - (- 1))}^{2} + {(8 - 7)}^{2}}$

$\implies \sf AP =\: \sqrt{{(4 + 1)}^{2} + {(1)}^{2}}$

$\implies \sf AP =\: \sqrt{{(5)}^{2} + {(1)}^{2}}$

$\implies \sf AP =\: \sqrt{5 \times 5 + 1 \times 1}$

$\implies \sf AP =\: \sqrt{25 + 1}$

$\implies \sf\bold{\red{AP =\: \sqrt{26}}}$

$\therefore$ The distance of AP is √26 .