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The points (-4, 0), (4, 0) and (0, 3) are the vertices of a

  1. right triangle

  2. isosceles triangle

  3. equilateral triangle

  4. scalene triangle

Coordinate Geometry

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Answer

Let A = (-4, 0), B = (4, 0) and C = (0, 3).

The points (-4, 0), (4, 0) and (0, 3) are the vertices of a? Coordinate Geometry, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

By distance formula,

d=(x2−x1)2+(y2−y1)2∴AB=[4−(−4)]2+(0−0)2=[4+4]2+02=82=64=8 units∴BC=(0−4)2+(3−0)2=(−4)2+32=16+9=25=5 units∴AC=[0−(−4)]2+(3−0)2=42+32=16+9=25=5 unitsd = \sqrt{(x2 - x1)^2 + (y2 - y1)^2} \\[1em] \therefore AB = \sqrt{[4 - (-4)]^2 + (0 - 0)^2} \\[1em] = \sqrt{[4 + 4]^2 + 0^2} \\[1em] = \sqrt{8^2} \\[1em] = \sqrt{64} \\[1em] = 8 \text{ units} \\[1em] \therefore BC = \sqrt{(0 - 4)^2 + (3 - 0)^2} \\[1em] = \sqrt{(-4)^2 + 3^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units} \\[1em] \therefore AC = \sqrt{[0 - (-4)]^2 + (3 - 0)^2} \\[1em] = \sqrt{4^2 + 3^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units} \\[1em]

Since, AC = BC.

∴ ABC is an isosceles triangle.

Hence, Option 2 is the correct option.

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