In the diagram below, P (1:0), Q (6:3) and R (9; −2) are the vertices of a triangle, such that PQ = QR and PQ…

Question

In the diagram below, P (1:0), Q (6:3) and R (9; −2) are the vertices of a triangle, such that PQ = QR and PQ ⊥ QR. T is the midpoint of PQ. S is the point of intersection of RQ and x-axis. V is a point on the x-axis such that ∠QSV = 121°. ∠QPS = θ

ScanSolve AI Tutor · Accepted Answer

Here's a step-by-step solution for Question 3:

3.1 Determine:

3.1.1 The length of PQ. Leave your answer in a surd form.

Use the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ with $P(1;0)$ and $Q(6;3)$ .
$PQ = \sqrt{(6-1)^2 + (3-0)^2}$ $PQ = \sqrt{(5)^2 + (3)^2}$ $PQ = \sqrt{25 + 9}$ $PQ = \boxed{\sqrt{34}}$
3.1.2 The gradient of PQ.
Use the gradient formula $m = \frac{y_2-y_1}{x_2-x_1}$ with $P(1;0)$ and $Q(6;3)$ .
$m_{PQ} = \frac{3-0}{6-1}$ $m_{PQ} = \boxed{\frac{3}{5}}$
3.1.3 The coordinates of T.
T is the midpoint of PQ.
Use the midpoint formula $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ with $P(1;0)$ and $Q(6;3)$ .
$T = \left(\frac{1+6}{2}, \frac{0+3}{2}\right)$ $T = \left(\frac{7}{2}, \frac{3}{2}\right)$ $T = \boxed{(3.5; 1.5)}$
3.2 Calculate:

3.2.1 The area of $\triangle QTR$ .
Given $PQ \perp QR$ , $\triangle PQR$ is a right-angled triangle at Q.
The area of $\triangle PQR = \frac{1}{2} \times PQ \times QR$ .
Given $PQ=QR$ and $PQ=\sqrt{34}$ (from 3.1.1), so $PQ^2 = 34$ .
$\text{Area of } \triangle PQR = \frac{1}{2} \times PQ^2 = \frac{1}{2} \times 34 = 17 \text{ square units}$
T is the midpoint of PQ, so $QT = \frac{1}{2} PQ$ .
$\triangle QTR$ shares the same height (QR) as $\triangle PQR$ relative to base PQ.
$\text{Area of } \triangle QTR = \frac{1}{2} \times QT \times QR = \frac{1}{2} \times \left(\frac{1}{2} PQ\right) \times QR$ $\text{Area of } \triangle QTR = \frac{1}{4} \times PQ \times QR = \frac{1}{4} \times PQ^2$ $\text{Area of } \triangle QTR = \frac{1}{4} \times 34 = \frac{17}{2}$ $\text{Area of } \triangle QTR = \boxed{8.5 \text{ square units}}$
3.2.2 The size of $\theta$ with reasons.
$\theta = \angle QPS$ .
Since P and S are on the x-axis, $\angle QPS$ is the angle of inclination of the line segment PQ with the positive x-axis.
The gradient of PQ is $m_{PQ} = \frac{3}{5}$ (from 3.1.2).
The angle of inclination $\theta$ is given by $\tan \theta = m_{PQ}$ .
$\tan \theta = \frac{3}{5}$ $\theta = \arctan\left(\frac{3}{5}\right)$ $\theta \approx 30.9637^{\circ}$ $\theta = \boxed{30.96^{\circ}}$
Reason: The angle $\theta$ is the angle of inclination of line PQ with the positive x-axis.

3.2.3 The coordinates of S.
S is the point of intersection of line RQ and the x-axis.
First, find the equation of line RQ using $Q(6;3)$ and $R(9;-2)$ .
Calculate the gradient of RQ.
$m_{RQ} = \frac{-2-3}{9-6} = \frac{-5}{3}$
Use the point-slope form $y - y_1 = m(x - x_1)$ with $Q(6;3)$ .
$y - 3 = -\frac{5}{3}(x - 6)$ $y - 3 = -\frac{5}{3}x + \frac{30}{3}$ $y - 3 = -\frac{5}{3}x + 10$ $y = -\frac{5}{3}x + 13$
To find S (on the x-axis), set $y=0$ .
$0 = -\frac{5}{3}x + 13$ $\frac{5}{3}x = 13$ $5x = 39$ $x = \frac{39}{5} = 7.8$ $S = \boxed{(7.8; 0)}$
Determine, with reasons, the gradient of a line through T and the midpoint of PR.
Let M be the midpoint of PR.
Use the midpoint formula with $P(1;0)$ and $R(9;-2)$ .
$M = \left(\frac{1+9}{2}, \frac{0+(-2)}{2}\right)$ $M = \left(\frac{10}{2}, \frac{-2}{2}\right)$ $M = (5; -1)$
T is the midpoint of PQ (from 3.1.3), $T=(3.5; 1.5)$ .
M is the midpoint of PR.
By the Midpoint Theorem, the line segment connecting the midpoints of two sides of a triangle is parallel to the third side.
Therefore, $TM \parallel QR$ .
Since $TM \parallel QR$ , their gradients are equal
$m_{TM} = m_{QR}$ . We already calculated $m_{QR}$ in 3.2.3.
$m_{TM} = m_{QR} = \boxed{-\frac{5}{3}}$
Reason: By the Midpoint Theorem, the line segment connecting the midpoints of two sides of a triangle is parallel to the third side. Since T is the midpoint of PQ and M is the midpoint of PR, TM is parallel to QR. Therefore, their gradients are equal.

In the diagram below, P (1:0), Q (6:3) and R (9; -2) are the vertices of a triangle, such that PQ = QR and PQ ⊥ QR. T is the midpoint of PQ. S is the point of intersection of RQ and x-axis. V is a point on the x-axis such that ∠QSV = 121°. ∠QPS = θ

Step-by-step solution

Use the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ with $P(1;0)$ and $Q(6;3)$ .

Use the gradient formula $m = \frac{y_2-y_1}{x_2-x_1}$ with $P(1;0)$ and $Q(6;3)$ .

T is the midpoint of PQ.

Given $PQ \perp QR$ , $\triangle PQR$ is a right-angled triangle at Q.

Given $PQ=QR$ and $PQ=\sqrt{34}$ (from 3.1.1), so $PQ^2 = 34$ .

T is the midpoint of PQ, so $QT = \frac{1}{2} PQ$ .

$\theta = \angle QPS$ .

The gradient of PQ is $m_{PQ} = \frac{3}{5}$ (from 3.1.2).

The angle of inclination $\theta$ is given by $\tan \theta = m_{PQ}$ .

S is the point of intersection of line RQ and the x-axis.

Calculate the gradient of RQ.

Use the point-slope form $y - y_1 = m(x - x_1)$ with $Q(6;3)$ .

To find S (on the x-axis), set $y=0$ .

Let M be the midpoint of PR.

T is the midpoint of PQ (from 3.1.3), $T=(3.5; 1.5)$ .

By the Midpoint Theorem, the line segment connecting the midpoints of two sides of a triangle is parallel to the third side.

Since $TM \parallel QR$ , their gradients are equal

Still stuck on this one?

More Mathematics Questions

In the diagram below, P (1:0), Q (6:3) and R (9; -2) are the vertices of a triangle, such that PQ = QR and PQ ⊥ QR. T is the midpoint of PQ. S is the point of intersection of RQ and x-axis. V is a point on the x-axis such that ∠QSV = 121°. ∠QPS = θ

Step-by-step solution

Use the distance formula d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}d=(x2​−x1​)2+(y2​−y1​)2​ with P(1;0)P(1;0)P(1;0) and Q(6;3)Q(6;3)Q(6;3).

Use the gradient formula m=y2−y1x2−x1m = \frac{y_2-y_1}{x_2-x_1}m=x2​−x1​y2​−y1​​ with P(1;0)P(1;0)P(1;0) and Q(6;3)Q(6;3)Q(6;3).

T is the midpoint of PQ.

Given PQ⊥QRPQ \perp QRPQ⊥QR, △PQR\triangle PQR△PQR is a right-angled triangle at Q.

Given PQ=QRPQ=QRPQ=QR and PQ=34PQ=\sqrt{34}PQ=34​ (from 3.1.1), so PQ2=34PQ^2 = 34PQ2=34.

T is the midpoint of PQ, so QT=12PQQT = \frac{1}{2} PQQT=21​PQ.

θ=∠QPS\theta = \angle QPSθ=∠QPS.

The gradient of PQ is mPQ=35m_{PQ} = \frac{3}{5}mPQ​=53​ (from 3.1.2).

The angle of inclination θ\thetaθ is given by tan⁡θ=mPQ\tan \theta = m_{PQ}tanθ=mPQ​.