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Mathematics

The coordinates of two points E and F are (0, 4) and (3, 7) respectively. Find :

(i) the gradient of EF.

(ii) the equation of EF.

(iii) the coordinates of the point where the line EF intersects the x-axis.

Straight Line Eq

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Answer

(i) Gradient of a line = y2y1x2x1\dfrac{y2 - y1}{x2 - x1}

Putting values in above formula we get,

Gradient =7430=33=1.\text{Gradient } = \dfrac{7 - 4}{3 - 0} \\[1em] = \dfrac{3}{3} \\[1em] = 1.

Hence, the gradient of EF is 1.

(ii) Equation of EF can be given by,

y - y1 = m(x - x1)

Putting values in above equation we get,

⇒ y - 4 = 1(x - 0)
⇒ y - 4 = x
⇒ x - y + 4 = 0.

Hence, the equation of EF is x - y + 4 = 0.

(iii) The coordinates where EF intersects x-axis will be where y = 0.

Substituting y = 0 in x - y + 4 = 0 ,

⇒ x - 0 + 4 = 0
⇒ x = -4.

Hence, coordinates where EF intersects x-axis are (-4, 0).

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