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Mathematics

Find the slope of the line passing through the points:

(i) A(–2, 1) and B(3, –4)

(ii) A(0, –3) and B(2, 1)

(iii) A(4, –9) and B(–2, –1)

(iv) A(2, 5) and B(–4, –4)

Straight Line Eq

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Answer

(i) A(-2, 1) and B(3, -4)

By formula,

Slope (m) = y2y1x2x1\dfrac{y2 - y1}{x2 - x1}

Substituting values we get,

m=413(2)=53+2=55=1.\Rightarrow m = \dfrac{-4 - 1}{3 - (-2)} \\[1em] = \dfrac{-5}{3 + 2} \\[1em] = \dfrac{-5}{5} \\[1em] = -1.

Hence, slope is -1.

(ii) A(0, –3) and B(2, 1)

By formula,

Slope (m) = y2y1x2x1\dfrac{y2 - y1}{x2 - x1}

Substituting values we get,

m=1(3)20=1+32=42=2.\Rightarrow m = \dfrac{1 - (-3)}{2 - 0} \\[1em] = \dfrac{1 + 3}{2} \\[1em] = \dfrac{4}{2} \\[1em] = 2.

Hence, slope is 2.

(iii) A(4, –9) and B(–2, –1)

By formula,

Slope (m) = y2y1x2x1\dfrac{y2 - y1}{x2 - x1}

Substituting values we get,

m=1(9)24=1+96=86=43.\Rightarrow m = \dfrac{-1 - (-9)}{-2 - 4} \\[1em] = \dfrac{-1 + 9}{-6} \\[1em] = \dfrac{8}{-6} \\[1em] = -\dfrac{4}{3}.

Hence, slope is 43-\dfrac{4}{3}.

(iv) A(2, 5) and B(–4, –4)

By formula,

Slope (m) = y2y1x2x1\dfrac{y2 - y1}{x2 - x1}

Substituting values we get,

m=4542=96=32.\Rightarrow m = \dfrac{-4 - 5}{-4 - 2} \\[1em] = \dfrac{-9}{-6} \\[1em] = \dfrac{3}{2}.

Hence, slope is 32\dfrac{3}{2}.

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