Question

COORDINATE GEOMETRY
CLASS 10
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•Please answer the given question in the picture.
i.e. question 7
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THANKS...​ ​

Answer 1

We've been given 4 vertices of a parallelogram, and we are asked to find the height of the line drawn perpendicular to AB from C.

In order to find the height, we'll find the area of ΔACB with the help of coordinates, then equate that area using the 1/2 b × h formula to find the height.

Coordinates:

C(-3, 2)

A(1, -2)

B(2, 3)

We know the area will be:

⇒  ar(ΔCAB) = ¹/₂ [x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)]

⇒  ar(ΔCAB) = ¹/₂ [-3(-2 - 3) + 1(3 - 2) + 2(2 - (-2))]

⇒  ar(ΔCAB) = ¹/₂ [-3(-5) + 1(1) + 2(2 + 2)]

⇒  ar(ΔCAB) = ¹/₂ [15 + 1 + 2(4)]

⇒  ar(ΔCAB) = ¹/₂ [15 + 1 + 8]

⇒  ar(ΔCAB) = ¹/₂ [16 + 8]

⇒  ar(ΔCAB) = ¹/₂ [24]

⇒  ar(ΔCAB) = 12 sq.units.

Now, we can also find the area of CAB by using 1/2 b × h since CE ⊥ AB

But first, we need to find the distance of AB.

$\sf \Longrightarrow AB = \sqrt{\Big\{x_1 - x_2\Big\}^2 + \Big\{y_1 - y_2\Big\}^2}$

$\sf \Longrightarrow AB = \sqrt{\Big\{1 - 2\Big\}^2 + \Big\{-2 - 3\Big\}^2}$

$\sf \Longrightarrow AB = \sqrt{\Big\{-1\Big\}^2 + \Big\{-5\Big\}^2}$

$\sf \Longrightarrow AB = \sqrt{1 + 25}$

$\sf \Longrightarrow AB = \sqrt{26} \ sq.units.$

Now, we find the height.

⇒ ar(ΔCAB) = ¹/₂ × b × h

Here the base is AB, and height is CE.

$\Longrightarrow \sf ar(\triangle CAB) = \dfrac{1}{2} \times b \times h$

$\sf \Longrightarrow 12 = \dfrac{1}{2} \times \sqrt{26} \times CE$

$\sf \Longrightarrow 12 \times 2 = CE\sqrt{26}$

$\sf \Longrightarrow 24 = CE\sqrt{26}$

$\sf \Longrightarrow CE\sqrt{26} = 24$

$\sf \Longrightarrow CE = \dfrac{24}{\sqrt{26}}$

$\sf \Longrightarrow CE = \dfrac{24}{\sqrt{26}} \times \dfrac{\sqrt{26}}{\sqrt{26}}$

$\sf \Longrightarrow CE =\dfrac{24\sqrt{26}}{\sqrt{26} \times \sqrt{26}}$

$\sf \Longrightarrow CE =\dfrac{24\sqrt{26}}{26}$

$\sf \Longrightarrow CE =\dfrac{12\sqrt{26}}{13} \ sq.units.$

Therefore the height of this parallelogram is (12√26)/13 sq.units.

Answer 2

Answer:

We've been given 4 vertices of a parallelogram, and we are asked to find the height of the line drawn perpendicular to AB from C.

In order to find the height, we'll find the area of ΔACB with the help of coordinates, then equate that area using the 1/2 b × h formula to find the height.

Coordinates:

C(-3, 2)

A(1, -2)

B(2, 3)

We know the area will be:

⇒ ar(ΔCAB) = ¹/₂ [x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)]

⇒ ar(ΔCAB) = ¹/₂ [-3(-2 - 3) + 1(3 - 2) + 2(2 - (-2))]

⇒ ar(ΔCAB) = ¹/₂ [-3(-5) + 1(1) + 2(2 + 2)]

⇒ ar(ΔCAB) = ¹/₂ [15 + 1 + 2(4)]

⇒ ar(ΔCAB) = ¹/₂ [15 + 1 + 8]

⇒ ar(ΔCAB) = ¹/₂ [16 + 8]

⇒ ar(ΔCAB) = ¹/₂ [24]

⇒ ar(ΔCAB) = 12 sq.units.

Now, we can also find the area of CAB by using 1/2 b × h since CE ⊥ AB

But first, we need to find the distance of AB.

\sf \Longrightarrow AB = \sqrt{\Big\{x_1 - x_2\Big\}^2 + \Big\{y_1 - y_2\Big\}^2}⟹AB=

{x

1

−x

2

}

2

+{y

1

−y

2

}

2

\sf \Longrightarrow AB = \sqrt{\Big\{1 - 2\Big\}^2 + \Big\{-2 - 3\Big\}^2}⟹AB=

{1−2}

2

+{−2−3}

2

\sf \Longrightarrow AB = \sqrt{\Big\{-1\Big\}^2 + \Big\{-5\Big\}^2}⟹AB=

{−1}

2

+{−5}

2

\sf \Longrightarrow AB = \sqrt{1 + 25}⟹AB=

1+25

\sf \Longrightarrow AB = \sqrt{26} \ sq.units.⟹AB=

26

sq.units.

Now, we find the height.

⇒ ar(ΔCAB) = ¹/₂ × b × h

Here the base is AB, and height is CE.

\Longrightarrow \sf ar(\triangle CAB) = \dfrac{1}{2} \times b \times h⟹ar(△CAB)=

2

1

×b×h

\sf \Longrightarrow 12 = \dfrac{1}{2} \times \sqrt{26} \times CE⟹12=

2

1

×

26

×CE

\sf \Longrightarrow 12 \times 2 = CE\sqrt{26}⟹12×2=CE

26

\sf \Longrightarrow 24 = CE\sqrt{26}⟹24=CE

26

\sf \Longrightarrow CE\sqrt{26} = 24⟹CE

26

=24

\sf \Longrightarrow CE = \dfrac{24}{\sqrt{26}}⟹CE=

26

24

\sf \Longrightarrow CE = \dfrac{24}{\sqrt{26}} \times \dfrac{\sqrt{26}}{\sqrt{26}}⟹CE=

26

24

×

26

26

\sf \Longrightarrow CE =\dfrac{24\sqrt{26}}{\sqrt{26} \times \sqrt{26}}⟹CE=

26

×

26

24

26

\sf \Longrightarrow CE =\dfrac{24\sqrt{26}}{26}⟹CE=

26

24

26

\sf \Longrightarrow CE =\dfrac{12\sqrt{26}}{13} \ sq.units.⟹CE=

13

12

26

sq.units.

Therefore the height of this parallelogram is (12√26)/13 sq.units.