The vertices of a triangle ABC are A (0, 5), B (-1, -2) and C (11, 7). Write down the equation of BC. Find :
(i) the equation of line through A and perpendicular to BC.
(ii) the co-ordinates of the point P, where the perpendicular through A, as obtained in (i), meets BC.

Question

The vertices of a triangle ABC are A (0, 5), B (-1, -2) and C (11, 7). Write down the equation of BC. Find :
(i) the equation of line through A and perpendicular to BC.
(ii) the co-ordinates of the point P, where the perpendicular through A, as obtained in (i), meets BC.

KnowledgeBoat · Accepted Answer

Then the equation of the line BC is ⇒ y - y₁ = m(x - x₁) ⇒ y - (-2) = [x - (-1)] ⇒ 4(y + 2) = 3(x + 1) ⇒ 4y + 8 = 3x + 3 ⇒ 3x - 4y = 8 - 3 ⇒ 3x - 4y = 5 ......... (1) (i) Let slope of line perpendicular to BC = m₁ As product of slope of perpendicular lines is -1 ∴ m x m₁ = -1 ⇒ x m₁ = -1 ⇒ m₁ = So, the required equation of the line through A (0, 5) and perpendicular to BC is given by ⇒ y - y₁ = m(x - x₁) ⇒ y - 5 = (x - 0) ⇒ 3y - 15 = -4x ⇒ 4x + 3y = 15 ......... (2) Hence, equation of the required line is 4x + 3y = 15. (ii) The required point P will be the point of intersection of lines (1) and (2). 3x - 4y = 5 ......... (1) 4x + 3y = 15 ......... (2) Multiplying equation (1) by 3 and (2) by 4 and adding we get, ⇒ 9x - 12y + 16x + 12y = 15 + 60 ⇒ 25x = 75 ⇒ x = ⇒ x = 3 Substituting the value of x in equation (1) we get, 3(3) - 4y = 5 ⇒ 4y = 3(3) – 5 ⇒ 4y = 9 - 5 ⇒ 4y = 4 ⇒ y = 1 Hence, the co-ordinates of the required point P is (3, 1).

Mathematics

The vertices of a triangle ABC are A (0, 5), B (-1, -2) and C (11, 7). Write down the equation of BC. Find :
(i) the equation of line through A and perpendicular to BC.
(ii) the co-ordinates of the point P, where the perpendicular through A, as obtained in (i), meets BC.

Straight Line Eq

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Mathematics

The vertices of a triangle ABC are A (0, 5), B (-1, -2) and C (11, 7). Write down the equation of BC. Find :(i) the equation of line through A and perpendicular to BC.(ii) the co-ordinates of the point P, where the perpendicular through A, as obtained in (i), meets BC.

Straight Line Eq

Related Questions

Find the value of k such that the line (k – 2)x + (k + 3)y – 5 = 0 is:(i) perpendicular to the line 2x – y + 7 = 0(ii) parallel to it.

A (8, -6), B (-4, 2) and C (0, -10) are vertices of a triangle ABC. If P is the mid-point of AB and Q is the mid-point of AC, use co-ordinate geometry to show that PQ is parallel to BC. Give a special name of quadrilateral PBCQ.

Given a straight line x cos 30° + y sin 30° = 2. Determine the equation of the other line which is parallel to it and passes through (4, 3).

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The vertices of a triangle ABC are A (0, 5), B (-1, -2) and C (11, 7). Write down the equation of BC. Find :
(i) the equation of line through A and perpendicular to BC.
(ii) the co-ordinates of the point P, where the perpendicular through A, as obtained in (i), meets BC.

Find the value of k such that the line (k – 2)x + (k + 3)y – 5 = 0 is:
(i) perpendicular to the line 2x – y + 7 = 0
(ii) parallel to it.

From the given figure, find:
(i) the co-ordinates of A, B and C.
(ii) the equation of the line through A and parallel to BC.