Question

If the line segment joining the points (7, -1) and (9, 3) be the base of an isosceles triangle and the abscissa of the vertex is 4, find its ordinate

Answer 1

☆ Given Question:-

• If the line segment joining the points (7, -1) and (9, 3) be the base of an isosceles triangle and the abscissa of the vertex is 4, find its ordinate.

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☆ To Find :-

• The ordinate of vertex.

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☆ Formula Used:-

☆ Distance Formula

Let us consider a line segment joining the points A and B,

then distance between A and B is given by

$\tt\implies \:AB = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

$\tt \: where \: A \: is \: (x_1,y_1) \: and \: B \: is \: (x_2,y_2).$

☆ In isosceles triangle, perpendicular drawn from the vertex bisects the opposite sides.

☆ Pythagoras Theorem

$\begin{gathered}\star\;{\boxed{\tt{{(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2}}}}\end{gathered}*$

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$\large\underline\purple{\bold{ \tt \: Solution :- }}$

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☆ Let us consider an isosceles triangle ABC such that AB = AC and let the coordinates of base BC be B(7, -1) and C(9, 3) respectively.

☆ Let coordinates of vertex A be (4, y).

☆ Construction:- Let us draw a perpendicular AD from

vertex A intersect base BC at D.

☆ So, that D is the midpoint of BC.

$\tt \: So, \: Coordinates \: of \: D \: = (\dfrac{9 + 7}{2} ,\dfrac{ - 1 + 3}{2} ) = (8, \: 1)$

$\tt \: Now, \: in \: right \: triangle \: ABD$

$\tt \: {AB}^{2} = {AD}^{2} + {BD}^{2}$

$\tt \: {(7 - 4)}^{2} + {(1 + y)}^{2} = {(8 - 4)}^{2} + {(1 - y)}^{2} + {(7 - 8)}^{2} + {( - 1 - 1)}^{2}$

$\tt \: 9 + 1 + {y}^{2} + 2y = 16 + 1 + {y}^{2} - 2y + 1 + 4$

$\tt \: { \cancel y}^{2} + 2y + 10 = { \cancel y}^{2} - 2y + 22$

$\tt\implies \:4y = 12$

$\tt\implies \:y \: = 3$

$\bf\implies \:Hence, \: ordinate \: of \: vertex \: is \: 3$