Algebra - Coordinate Geometry: y = mx + c

Identify the gradient and y-intercept of the line with the equation $y = -4x + 7$.

Gradient ($m$) = -4, y-intercept ($c$) = 7.

Compare the given equation to the standard form $y = mx + c$. Here, the coefficient of $x$ is -4 and the constant term is 7.

Find the gradient of the line passing through the points $A(2, 3)$ and $B(5, 12)$.

$m = 3$

Using the gradient formula $m = \frac{y_2 - y_1}{x_2 - x_1}$, we get $m = \frac{12 - 3}{5 - 2} = \frac{9}{3} = 3$.

Rearrange the equation $3x + 2y = 8$ into the form $y = mx + c$ and state the gradient.

$y = -\frac{3}{2}x + 4$; Gradient = -1.5

Subtract $3x$ from both sides to get $2y = -3x + 8$. Divide every term by 2 to isolate $y$: $y = -\frac{3}{2}x + 4$.

Find the equation of a line that has a gradient of 2 and passes through the point $(3, 10)$.

$y = 2x + 4$

Substitute $m = 2$, $x = 3$, and $y = 10$ into $y = mx + c$: $10 = 2(3) + c$. This simplifies to $10 = 6 + c$, so $c = 4$. Plug $m$ and $c$ back into the general form.

Determine if the lines $y = 5x - 2$ and $10x - 2y = 4$ are parallel.

Yes, they are parallel.

The first line has a gradient of 5. Rearranging the second line: $-2y = -10x + 4 \Rightarrow y = 5x - 2$. Since both lines have the same gradient ($m=5$), they are parallel.

$y = mx + c$ $y = 3x - 2$ To find the $x$-intercept, set $y = 0$: $0 = 3x - 2$ $2 = 3x$ $x = \frac{2}{3}$

The value $c$ is the $y$-intercept, which is given as $-2$ (the $y$-coordinate when $x=0$). The gradient $m$ is given as $3$. To find where the line crosses the $x$-axis, we solve for $x$ when $y$ is zero.

First, find the gradient $m$: $m = \frac{1 - 4}{4 - (-2)} = \frac{-3}{6} = -\frac{1}{2}$ Now use $y = mx + c$ with point $(4, 1)$: $1 = -\frac{1}{2}(4) + c$ $1 = -2 + c$ $c = 3$ Equation: $y = -\frac{1}{2}x + 3$

We calculate the gradient using the change in $y$ divided by the change in $x$. Once we have $m$, we substitute one of the points into the general equation to solve for the $y$-intercept $c$.