Question

(c) PQ is a straight line of 13 units If P has the co-ordinates (5, - 3) and Q has the co-ordinates
(-7. v): find the values of y'.

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Answer 1

Given :

•Distance between PQ =13 units

•P=(5,-3)

•Q=(-7,v)

To Find :

• The value of y=?

Formula :

•Distance between two points is

$\\ \red\bigstar\boxed{\sf{AB=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}}$

Solution :

• Distance between PQ is 13 units

• P=(5,-3)

• Q=(-7,v)

By using Distance Formula

$\\ \implies\sf{PQ=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}$

Substituting the given values :

$\\ \implies\sf{PQ=\sqrt{(v-(-3))^{2}+(-7-5)^{2}}}$

$\\ \implies\sf{13= \sqrt{(v+3)^{2}+(-12)^{2}}}$

$\\ \implies\sf{169=(v+3)^{2}+144}$

$\\ \implies\sf{169=v^{2}+9+6v+144}$

$\\ \implies\sf{169-144-v^{2}-9-6v=0}$

$\\ \implies\sf{169-153-v^{2}-6v=0}$

$\\ \implies\sf{16-v^{2}-6v=0}$

$\\ \implies\sf{-v^{2}-6v+16=0}$

$\\ \implies\sf{v^{2}+6v-16=0}$

$\\ \implies\sf{v^{2}+8v-2v-16=0}$

$\\ \implies\sf{v(v+8)-2(v+8)=0}$

$\\ \implies\sf{(v+8)(v-2)=0}$

$\\ \implies \boxed { \sf{v = - 8}} \sf \: and \: \boxed{ \sf{v = 2}}$

➡ The value of v is 2