Write down the gradient of L₁ .

Line L₂, is perpendicular to line L₁, and passes through the point (4, 5) .

(i) Write down the gradient of L₂ .

(ii) Find the equation of L₂ .

Line L₂, is perpendicular to line L₁, and passes through the point (4, 5) .

Write down the coordinates of the point of intersection of L₁ and L₂ .

–2     (A1)     (C1)

Note: Do not accept \(\frac{{ - 2}}{1}\)

[1 mark]

(i) \(\frac{1}{2}(0.5)\)     (A1)(ft)

Note: Follow through from their part (a).

(ii) \(5 = \frac{1}{2}(4) + c\)     (M1)

Note: Award (M1) for their gradient substituted correctly.

\(y = \frac{1}{2}x + 3\)     (A1)(ft)

Note: Follow through from their part (b)(i).

OR

\(y - 5 = \frac{1}{2}(x - 4)\)     (M1)(A1)(ft)     (C3)

Notes: Award (M1) for their gradient substituted correctly, (A1)(ft) for 5 and 4 seen in the correct places. Follow through from their part (b)(i).

[3 marks]

(0.8, 3.4) or \(\left( {\frac{4}{5},\frac{{17}}{5}} \right)\)     (A1)(ft)(A1)(ft)     (C2)

Notes: Accept x = 0.8 and y = 3.4. Award (A1)(ft) for an attempt to solve the equations analytically, (attempt to eliminate either x or y), or graphically with a sketch (two reasonably accurate straight line graphs (from their answer to part (b)) and an indication of scale). Follow through from their L₂ if it intersects L₁, OR follow through from their equation, or expression in x, from their part (b)(ii). Award at most (A1)(ft)(A0)(ft) if brackets missing. Award (A0)(ft)(A1)(ft) for an answer of (0, 5) following an equation (or expression in x) of the form y = mx + 5 (m ≠ –2) found in part (b).

[2 marks]

The majority of candidates were able to write down the gradient of the straight line in part (a) but a correct answer for the gradient of the perpendicular proved to be more elusive in part (b)(i). Many however recovered in the remainder of the question as they were able to find the equation of a line using their gradient and the coordinates of a point on the line but, in some case, did not always show clear working. A significant minority of candidates, who attempted to substitute (4, 5) into the equation y = mx + c , incorrectly identified the value of c as 5.

The majority of candidates were able to write down the gradient of the straight line in part (a) but a correct answer for the gradient of the perpendicular proved to be more elusive in part (b)(i). Many however recovered in the remainder of the question as they were able to find the equation of a line using their gradient and the coordinates of a point on the line but, in some case, did not always show clear working. A significant minority of candidates, who attempted to substitute (4, 5) into the equation y = mx + c , incorrectly identified the value of c as 5.

The majority of candidates were able to write down the gradient of the straight line in part (a) but a correct answer for the gradient of the perpendicular proved to be more elusive in part (b)(i). Many however recovered in the remainder of the question as they were able to find the equation of a line using their gradient and the coordinates of a point on the line but, in some case, did not always show clear working. A significant minority of candidates, who attempted to substitute (4, 5) into the equation y = mx + c , incorrectly identified the value of c as 5.