straight line, the question is- Find where the line joining the points $(-3,5)$ and $(-4,8)$ meets the line $x=15$. the question is- Find where the line joining the points $(-3,5)$ 
and $(-4,8)$ meets the line $x=15$ .
all i could do was work out the gradient $-3$ 
please help!!
im not sure how to work this or if i did it right
 A: To expand on Sandro Lovinički’s first answer, the gradient, a.k.a. slope, of the line is $$m = {y_2 - y_1 \over x_2-x_1}.$$ This is an invariant of the line: it is the same no matter which two pairs of points you choose. You therefore have $${8-5 \over (-4)-(-3)} = {y-5 \over 15-(-3)}.$$ Solve for $y$.  
This approach essentially uses proportions derived from similar triangles.
A: Hint : Write the equation of st. line joining the points $(-3,5)$ and $(-4,8)$. Put $x=15$ in the equation. You'll get the value of y-coordinate.
A: Gradient approach
What is the meaning of gradient $-3$ you got? Gradient is the slope of the line; it tells us for how many units $y$ changes (in your case drops by $3$) when $x$ increases by $1$. Only from this you can conclude what $y$ will be when $x$ is $15$ (when it "meets the line $x=15$") because you know what $y$ was at $x$ values of $-3$ and $-4$ (only one is enough for this deduction).
"Standard approach"
First we need to find the line equation connecting points $(x_1,y_1)=(-3,5)$ and $(x_2,y_2)=(-4,8)$. We know that a basic form of a line equation is
\begin{equation}
y = a \cdot x + b
\end{equation}
where $a,b \in \mathbb{R}$ are values we need to find out. We are looking for a line on which both of our points are, so we want our two points' $x$ and $y$ coordinates satisfy the above general line equation. If we write that which we just stated (what we want), we get
\begin{align*}
y_1 &= a \cdot x_1 + b\\
y_2 &= a \cdot x_2 + b
\end{align*}
where only $a,b$ are unknowns when we input our values of $x_1,y_1,x_2,y_2$. You should try to solve this yourself.
When you get the solution, i.e values for $a$ and $b$, pluck them into $y=a \cdot x + b$ and that is the equation of a line passing through your $2$ points.
The the task directs you to find the intersection of that line and line $x=15$ which maybe seems confusing because it is not of the form $y=ax+b$. It just mean that there is no relationship between $x$ and $y$, that $x$ is always $15$ and $y$ is arbitrary. You can visualize or draw that; it is the line that is parallel to the $y$-axis and passes through $15$ at $x$-axis.
To find an intersection of two lines
\begin{align*}
y &= a_1 \cdot x + b_1\\
y &= a_2 \cdot x + b_2,
\end{align*}
means that we have to find a point $(x_0,y_0)$ that satisfies both of them because that point is the only point on both of the lines. Generally, to find an intersection, we solve the above system of equations, but in our case it is a simpler system because we have
\begin{align*}
y &= a \cdot x + b \quad \text{(line through your two points in the task)}\\
x &= 15 \quad \text{(second line)},
\end{align*}
Therefore, we know that the intersection $(x_0,y_0)$ we are looking for will have $x_0 = 15$ (remember, the intersection must satisfy both line equations, so to satisfy the second one it must have 15 on $x$-coordinate). Now when we know $x_0$, we calculate $y_0$ using the first equation
\begin{equation}
y_0 = a \cdot 15 + b
\end{equation}
and because you already know $a,b$ from one of the first steps I explained (finding line equation passing through your two points), you directly calculate $y_0$.
There, you have your solution $(x_0,y_0)$. Feel free to share what you got in the comments.
A: Note: for advanced discussion of this, please refer to: Wiki-Line–line intersection.
When $2$ lines intersect, the point of intersection satisfies the equation of each of the intersecting lines. In your case, you have one equation fully written as $x=15$. However the other equation is not known. 
You want to find the equation for a line that passes through the two points:
$(-3,5) and (-4,8)$.
it is common to work with points using (x,y) coordinates. So we have:
point 1 having $(x1,y1)$ as $(-3,5) ==> x1=-3$ and $y1=5$
similarly for point 2, we have $x2=-4$ and $y2=8$
The line equation can be written in this form:
$y = mx+b$ Where:
m is the slope value, and
b is the y-intercept value
The slope of the above line, given $2$ points can be calculated as:
$m=(y2-y1)/(x2-x1)$
given the $2$ points we have, m can be found as:
m=$(8-5)/(-4-(-3))=3/-1=-3$
So, the line equation is now:
$y=-3x+b$
But b is not known yet. We can find b by using the values from any of the given points. Lets use the value of point $1$, we know that when $x=-3$, $y=5$ because the point $1$ has the coordinates $(-3,5)$. This means that when we substitute the values in the above equation we get:
$5=-3(-3)+b$
From the above we can determine $b=5-9=-4$
Now we can write the complete line equation after we have determined m and b as:
$y=-3x+(-4)==> y=-3x-4$
When two lines intersect the x value will be the same on each line.
We need to re-write the above equation in terms of $x=...$ so we can use this fact.
$y=-3x-4 ==> y+4=-3x ==> x= (y+4)/(-3)$
At the intersection point the x-values are the same, so:
$(x=15)=(y+4)/-3$
That is:
$(15)*(-3)=y+4 ==> y=(15*-3)-4 =-49 ==> y=-49$
Since the 2 lines intersect, the x-values and y-values are the same for both lines. We have determined that $y=-49$ and we were given that $x=15$, so the point (15,-49) is the point of intersection of the 2 lines.
Note that this point, satisfies the equation $x=15$ and also satisfies the line equation we have constructed $y=-3x-4$ which you can test by substituting x=15 to get y=-49 as expected.
To verify the answer, it is sometimes possible to plot the two lines as follows:

A: Without explaining why, I present the following steps:

*

*Line connecting the two points $[-3,5]$ and $[-4,8]$ is described in terms of $ax+by+c=0$ coordinates $(a,b,c)$ as
$$ \pmatrix{a \\ b \\ c} = \left[ \matrix{-3 \\ 5 \\ 1} \right] \times \left[ \matrix{-4 \\ 8 \\ 1} \right] = \pmatrix{-3 \\ -1 \\ -4} $$


*Line described by $x=15$ seen as connecting two points $[15,0]$ and $[15,1]$ with $(a',b',c')$ coordinates
$$ \pmatrix{a' \\b' \\c'} = \left[ \matrix{15 \\ 0 \\ 1} \right] \times \left[ \matrix{15 \\ 1 \\ 1} \right] = \pmatrix{-1 \\ 0 \\ 15} $$


*Point joining the two lines in homogeneous coordinates $(x \,w, y \,w, w)$
$$ \left[ \matrix{x\,w \\ y \,w \\ w} \right] = \pmatrix{-1 \\ 0 \\ 15} \times \pmatrix{-3 \\ -1 \\ -4} = \left[ \matrix{15 \\ -49 \\ 1} \right] $$ The joining point has cordinates $x=\frac{15}{1} = 15$ and $y=\frac{-49}{1}=-49$.
NOTE: $\times$ here is the vector cross product.
