Find the area of the triangle whose vertices are (-3,5), (5,4) and (5,

1.	Find the area of the triangle whose vertices are (-3,5), (5,4) and (5,-2).
Answer» $\large\underline{\sf{Solution-}}$ The GIVEN vertices of triangle are ( - 3, 5 ), ( 5, 4 ) and ( 5, - 2 ). Now, we know that to find the area of triangle having vertices, the formula is given by $\boxed{ \rm{\ Area =\dfrac{1}{2} [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]}}$ Here, Coordinates are ( - 3, 5 ), ( 5, 4 ) and ( 5, - 2 ). So, • x₁ = - 3 • x₂ = 5 • x₃ = 5 • y₁ = 5 • y₂ = 4 • y₃ = - 2 So, on substituting the values, we get $\rm :\longmapsto\:Area =\dfrac{1}{2} \| x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \|$ $\rm \: = \: \: \dfrac{1}{2} \| - 3(4 - ( - 2)) + 5( - 2 - 5) + 5(5 - 4)\|$ $\rm \: = \: \: \dfrac{1}{2} \| - 3(4 + 2) + 5( - 7) + 5(1)\|$ $\rm \: = \: \: \dfrac{1}{2} \| - 3(6) - 35 + 5\|$ $\rm \: = \: \: \dfrac{1}{2} \| - 18 - 30\|$ $\rm \: = \: \: \dfrac{1}{2} \| - 48\|$ $\rm \: = \: \: \dfrac{1}{2} \|48\|$ $\rm \: = \: \: \ \: 24 \: square \: units$ Additional Information :- Distance Formula :- • Distance formula is used to find the distance between two given Points. $\bf\implies \:AB = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$ Section Formula:- • Section Formula is used to find the co ordinates of the POINT(Q) which divides the line segment joining the points A and B internally in the ratio m : n ${\underline{\boxed{\rm{\quad \Big(x, y \Big) = \Bigg(\dfrac{mx_2 + nx_1}{m + n}, \dfrac{my_2 + ny_1}{m + n}\Bigg) \quad}}}}$ Midpoint Formula :- • Mid Point formula is used to find the Mid point on any line segment. ${\underline{\boxed{\rm{\quad \Big(x, y \Big) = \Bigg(\dfrac{x_2 + x_1}{2}, \dfrac{y_2 + y_1}{2}\Bigg) \quad}}}}$

Find the area of the triangle whose vertices are (-3,5), (5,4) and (5,-2).

Answer»

$\large\underline{\sf{Solution-}}$

The GIVEN vertices of triangle are ( - 3, 5 ), ( 5, 4 ) and ( 5, - 2 ).

Now, we know that to find the area of triangle having vertices, the formula is given by

$\boxed{ \rm{\ Area =\dfrac{1}{2} [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]}}$

Here, Coordinates are ( - 3, 5 ), ( 5, 4 ) and ( 5, - 2 ).

So,

• x₁ = - 3

• x₂ = 5

• x₃ = 5

• y₁ = 5

• y₂ = 4

• y₃ = - 2

So, on substituting the values, we get

$\rm :\longmapsto\:Area =\dfrac{1}{2} | x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) |$

$\rm \: = \: \: \dfrac{1}{2} | - 3(4 - ( - 2)) + 5( - 2 - 5) + 5(5 - 4)|$

$\rm \: = \: \: \dfrac{1}{2} | - 3(4 + 2) + 5( - 7) + 5(1)|$

$\rm \: = \: \: \dfrac{1}{2} | - 3(6) - 35 + 5|$

$\rm \: = \: \: \dfrac{1}{2} | - 18 - 30|$

$\rm \: = \: \: \dfrac{1}{2} | - 48|$

$\rm \: = \: \: \dfrac{1}{2} |48|$

$\rm \: = \: \: \ \: 24 \: square \: units$

Additional Information :-

Distance Formula :-

• Distance formula is used to find the distance between two given Points.

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

Section Formula:-

• Section Formula is used to find the co ordinates of the POINT(Q) which divides the line segment joining the points A and B internally in the ratio m : n

${\underline{\boxed{\rm{\quad \Big(x, y \Big) = \Bigg(\dfrac{mx_2 + nx_1}{m + n}, \dfrac{my_2 + ny_1}{m + n}\Bigg) \quad}}}}$

Midpoint Formula :-

• Mid Point formula is used to find the Mid point on any line segment.

${\underline{\boxed{\rm{\quad \Big(x, y \Big) = \Bigg(\dfrac{x_2 + x_1}{2}, \dfrac{y_2 + y_1}{2}\Bigg) \quad}}}}$

Find the area of the triangle whose vertices are (-3,5), (5,4) and (5,-2).

Additional Information :-

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