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Mathematics

The co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively. Find :

(i) the gradient of PQ;

(ii) the equation of PQ;

(iii) the co-ordinates of the point where PQ intersects the x-axis.

Straight Line Eq

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Answer

(i) By formula,

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

Substituting values we get,

Slope of PQ =5632=15=15.\text{Slope of PQ } = \dfrac{5 - 6}{-3 - 2} \\[1em] = \dfrac{-1}{-5} \\[1em] = \dfrac{1}{5}.

Hence, gradient of slope PQ = 15\dfrac{1}{5}.

(ii) By point slope form,

Equation : y - y1 = m(x - x1)

⇒ y - 6 = 15\dfrac{1}{5}(x - 2)

⇒ 5(y - 6) = x - 2

⇒ 5y - 30 = x - 2

⇒ 5y = x - 2 + 30

⇒ 5y = x + 28

Hence, equation of PQ is 5y = x + 28.

(iii) The point where PQ intersects x-axis, there y co-ordinate = 0.

Substituting y = 0 in equation of PQ we get,

⇒ 5 × 0 = x + 28

⇒ 0 = x + 28

⇒ x = -28.

Point = (x, y) = (-28, 0).

Hence, co-ordinates of the point where PQ intersects the x-axis = (-28, 0).

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