मराठी

Draw the graphs of 3x = 4y + 32 and 3x + 4y = 16. Find the coordinates of the vertices of the triangle formed by the lines with y + 2 = 0. Find the perimeter of the triangle. - Mathematics

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प्रश्न

Draw the graphs of 3x = 4y + 32 and 3x + 4y = 16. Find the coordinates of the vertices of the triangle formed by the lines with y + 2 = 0. Find the perimeter of the triangle.

बेरीज
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उत्तर

Step 1: Convert the equations into slope-intercept form

line 1:

3x = 4y + 32 ⇒ 4y = 3x − 32 ⇒ y = `3/4`x − 8

line 2:

3x + 4y = 16 ⇒ 4y = −3x + 16 ⇒ y = `− 3/4` x + 4

line 3:

y + 2 = 0 y = −2

Step 2: Find the coordinates of the vertices of the triangle

Vertex A:

Put x = 0 in y = `−3/4x` + 4:

y = 4 ⇒ Point A is (0, 4)

Vertex B:

Put x = 0 in y = `3/4x` − 8:

y = −8. Point B is (0, −8)

Vertex C:

Put y = −2 into Line 1:

`−2 = 3/4x − 8`

`= 3/4x = 6`

x = 8

So, point C = (8, −2)

A = (0, 4), B = (0, −8), C = (8, −2) ... [Vertices of the triangle]

Step 3: Use the distance formula to find side lengths.

The distance formula between two points (x1, y1) and (x2, y2) is:

`"Distance" = sqrt((x_2 − x_1)^2 + (y_2 − y_1)^2)`

For AB:

AB = `sqrt((0 − 0)^2 + (4 − (−8))^2)​`

= `sqrt(0 + 144)​`

= 12

For BC:

BC = `sqrt((8 − 0)^2 + (−2 − (−8))^2)`

​= `sqrt(64 + 36)`​

= `sqrt100​`

 =10

For CA:

CA = `sqrt((8 − 0)^2 + (−2 − 4)^2)`

​= `sqrt(64+36)`

​= `sqrt100`

​=10

Step 4: Find the perimeter

Perimeter = AB + BC + CA

= 12 + 10 + 10

= 32 units

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पाठ 21: Coordinate Geometry - EXERCISE 21B [पृष्ठ २५७]

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बी निर्मला शास्त्री Mathematics [English] Class 9 ICSE
पाठ 21 Coordinate Geometry
EXERCISE 21B | Q 3. (iv) | पृष्ठ २५७
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