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In the diagram, R and A are the x- and y-intercepts respectively of the straight line AR. The equation of AR is $y= -\dfrac{1}{2}x +4$. Another straight line cuts the y-axis at P(0;2) and passes through the points M(k;0) and N(3;4).

$\alpha$ and $\beta$ are the angles of inclination of the lines MN and AR respectively.


  1. Given that M, P and N are collinear points, calculate the value of k
  2. Determine the size of $\theta$, the obtuse angle between the two lines.
  3. Calculate the length of MR
  4. Calculate the area of $\bigtriangleup{MNR}$


in Mathematics by Diamond (79,336 points) | 277 views

1 Answer

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Best answer

1. grad MP= gradPN




Given 2 points on any line, we can determine the equation of the line in the form $$y=mx+c$$ where

$m$ is the gradient

$c$ is the $y$-intercept

To calculate the gradient, $m$ between $P$ and $N$, we use the formula:

\[ \dfrac{y_2-y_1}{x_2-x_1}\]


Since $c$ is the $y$-intercept i.e. the pont on the line when $x=0$ and in this case the point has the coordinates $P(0;2)$ thus $c=2$

$\therefore y=\dfrac{2}{3}x+2$

We are told that the 3 points $MP$ and $N$ are collinear, so it implies that the lines $MP$ and $PN$ have the same gradient, so:


$\therefore  k=-3$



$= 119.74^{\circ}$



Given the gradients of the two straight lines we can calculate the values of $\alpha$ and $\beta$.

To determine the value of $\alpha$ we use the fact that have a right-angle formed between the two axes, thus

\[\tan{\alpha}=\dfrac{O}{A} = \dfrac{2}{3} \]

which is just the gradient that we calculated in the first question, so

\[\alpha = \tan^{-1}\left({\frac{2}{3}}\right) = 33.69^{\circ}\]


Using the same thinking, we can infer that 

\[\tan{\beta}=\dfrac{O}{A} = \dfrac{-1}{2} \]


\[\beta = \tan^{-1}\left({\frac{-1}{2}}\right) = -26.57^{\circ}+180^{\circ} = 153.43^{\circ}   \]


$$\theta = \beta - \alpha $$

$$= 153.43^{\circ} - 33.69^{\circ} $$

$$= 119.74^{\circ}$$



11 units


We know that the coordinates for the point $M$ are $(-3,0)$, if we determine the coordinates of $R$ then we will be able to get the length of $MR$.

Since $R$ is on the line $y= \frac{-1}{2}x+4$, and knowing that the value of $y$ at that point is $0$, i.e.

\[ 0  = \frac{-1}{2}x+4 \]

Solving for $x$ gives us $8$, thus $R(8;0)$

Now we can get the length of the the line $MR$

\[x_R - x_M = 8 - (-3) = 11 \]



22 square units


Remember that the area of a triangle is given by the formula

\[ A = \dfrac{1}{2} \times B \times H \]

where $B$ is the base and $H$ is the height of the triangle. (Remember that the base and the height of any triangle are always at $90^{\circ}$ to each other)

\[ A_{\triangle{MNR}} = \dfrac{1}{2} \times 11 \times 4 = 22 \]

by Diamond (44,112 points)
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1 0
Please explain more how you got Tan b= -1/2
1 0
The line AR has gradient(m-coefficient of x) of -1/2.

Hence we know tan b (betta) is opposite/adjacent (y/x) which is also gradient value.
1 0
How did you get line AR gradient to be -1/2? What values of Y2,Y1,X2 and X1 did you use to arrive at that assuming you did it Y2-Y1/X2-X1
0 0
The gradient of $\dfrac{1}{2}$ is given and was not caluculated, it part of the question.

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