Find the equation of the side BC of the triangle ABC whose vertices are A (-1, -2), B (0, 1) and C (2, 0) respectively. Also, find the equation of the median through A (-1, -2).

Asked by Aaryan | 1 year ago |  36

Solution :-

Given: A line which is perpendicular and parallel to x–axis respectively and passing through (4, 3)

By using the formula,

The equation of line: [y – y1 = m(x – x1)]

Let us consider,

Case 1: When Line is parallel to x–axis

The parallel lines have equal slopes,

And, the slope of x–axis is always 0, then

The slope of line, m = 0

Coordinates of line are (x1, y1) = (4, 3)

The equation of line is y – y1 = m(x – x1)

Now substitute the values, we get

y – (3) = 0(x – 4)

y – 3 = 0

Case 2: When line is perpendicular to x–axis

The line is perpendicular to the x–axis, then x is 0 and y is – 1.

The slope of the line is, m = y/x

= -1/0

Coordinates of line are (x1, y1) = (4, 3)

The equation of line = y – y1 = m(x – x1)

Now substitute the values, we get

y – 3 = ($$\dfrac{-1}{0}$$) (x – 4)

x = 4

The equation of line when it is parallel to x – axis is y = 3 and it is perpendicular is x = 4.

Answered by Aaryan | 1 year ago

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