Slope and y-intercept

In the form y = mx + c, the coefficient of x, m, is the slope. The slopes of the first and second lines are q - 2 and 5 respectively. For equal slopes, q - 2 = 5, so q = 7. If q were 6, the first slope would be 4, not 5. Exam tip: for parallel lines or lines with equal slopes, equate their coefficients of x.

In the form \(y=mx+c\), the \(y\)-intercept is the constant term \(c\). The \(y\)-intercept of \(y=-6x+s\) is \(s\), while that of \(y=2x-9\) is \(-9\). Since the intercepts are equal, \(s=-9\). Note that \(-6\) is the slope of the first line, not its intercept. Exam tip: put \(x=0\) to find the \(y\)-intercept.

In the form y=mx+c, the slope is the coefficient of x, m, and the y-intercept is the constant term, c. Here, the slope is a+3 and the y-intercept is a. Therefore, their difference is (a+3)-a=3, which remains true for every value of a. Although a=3 and a=0 also give a difference of 3, the statement is not restricted to those values. Exam tip: In y=mx+c, first identify the coefficient of x and then the constant term.

In \(y=mx+c\), \(c\) is the y-intercept and \(m\) is the slope. In option A, both intercepts are \(-3\), while the slopes are \(2\) and \(-1\). In B, equal slopes make the lines parallel. Exam tip: compare \(c\) first, then \(m\).

The given line has slope \(\frac{3}{2}\), so a perpendicular line has slope \(-\frac{2}{3}\). With y-intercept 4, A follows. B is parallel. Tip: perpendicular slopes multiply to \(-1\).

Write the equation in \(y=mx+c\) form: \(2y=-3x+8\), so \(y=-\frac{3}{2}x+4\). Hence the slope is \(-\frac{3}{2}\) and the y-intercept is 4. Exam tip: check signs while transposing terms.

In \(y=0x-6\), the coefficient of x is 0, so the slope is 0 and the line is horizontal, parallel to the x-axis. Putting \(x=0\) gives \(y=-6\), so the y-intercept is \(-6\). Exam tip: in \(y=mx+c\), \(m=0\) always represents a horizontal line.

In \(y=mx+c\), \(c\) is the y-intercept and \(m\) is the slope. In option A, both lines have \(c=5\), while their slopes are 2 and -3. Exam tip: compare the x-coefficient and constant term separately.

The slope is \(\frac{7}{3}\), meaning that an increase of 3 units in \(x\) increases \(y\) by 7 units. For \(\Delta y=21\), we use \(\frac{7}{3}\Delta x=21\). Hence, \(\Delta x=21\times\frac{3}{7}=9\). If \(x\) increased by 7, then \(y\) would increase by only \(\frac{49}{3}\), not 21. Exam tip: for a line \(y=mx+c\), use \(\Delta y=m\Delta x\) in change-based questions.

In y = px + q, p is the slope of the line and q is its y-intercept. Here, p = -5, so y decreases as x increases and the line falls from left to right. Since q = -4, the line meets the y-axis at -4, giving a negative y-intercept. Option C is incorrect because a horizontal line has slope 0. Exam tip: the sign of the slope shows whether a line rises or falls, while the constant term gives the y-intercept.

\(x=5\) is a vertical line and never meets the y-axis, whose equation is \(x=0\); hence it has no y-intercept. In contrast, \(y=2x+1\) meets it at 1. Exam tip: \(x=c\), for \(c\ne0\), has no y-intercept.

In \(x=5\), the x-coordinate is fixed, so the graph is a vertical line parallel to the y-axis. Since the y-axis is \(x=0\), it cannot meet \(x=5\). Exam tip: any equation \(x=\text{constant}\) represents a vertical line.

A line in the form \(y=mx+c\) has \(c\) as its \(y\)-intercept. Here, \(2n-5=7\). Thus, \(2n=12\) and \(n=6\). Therefore, the slope is \(n-3=6-3=3\). Option 2 is not the slope; it does not result from substituting the value of \(n\). Exam tip: in \(y=mx+c\), the coefficient of \(x\) is always the slope.

In \(y=mx+c\), \(m\) is the slope and \(c\) is the y-intercept. In option A, \(m=-3\) is negative and \(c=5\) is positive. Option C has a negative slope but a negative intercept. Exam tip: check the signs of both coefficients separately.

The slope is \(\frac{\Delta y}{\Delta x}=\frac{18}{3}=6\). The point \((0,-8)\) shows that the \(y\)-intercept is \(-8\). Therefore, using slope-intercept form \(y=mx+c\), the line is \(y=6x-8\). In \(y=18x-8\), the intercept is correct, but the slope is 18, not 6. Exam tip: find the slope from the changes first, then use the point with \(x=0\) to identify the \(y\)-intercept.

When \(x\) increases by \(4\) and \(y\) decreases by \(20\), the slope is \(m=\frac{-20}{4}=-5\). Since the line meets the \(y\)-axis at \((0,12)\), its \(y\)-intercept is \(12\). Therefore, the equation is \(y=-5x+12\). Although \(y=-20x+12\) has the correct intercept, its slope is \(-20\), not \(-5\). Exam tip: Use a negative sign for the change in \(y\) when the quantity decreases.

In \(y=mx+c\), \(m>0\) means the line rises left to right, and \(c=-5\) puts its y-intercept below zero. Thus A is correct; B has \(m<0\). Exam tip: inspect \(m\) and \(c\) separately.

In the form y = mx + c, the y-intercept is c. Here, the y-intercept is u + 3, so u + 3 = 11 gives u = 8. Therefore, the slope is 2u - 1 = 2(8) - 1 = 15. Option 16 can result from forgetting to subtract 1. Exam tip: first find the parameter using the y-intercept, then substitute it into the coefficient of x.

Use the slope-intercept form y = mx + c. From 2x − 3y + 9 = 0, move the other terms to isolate the y-term: −3y = −2x − 9. Dividing by −3 gives y = (2/3)x + 3. Thus the coefficient of x is the slope m = 2/3, and the constant c = 3 is the y-intercept. Hence option A is correct. Option B results from losing the sign while dividing, whereas options C and D confuse the original coefficients or constant with the values in slope-intercept form. Substitution also checks the result: when x = 0, y = 3, satisfying the original equation.

To find a line’s y-intercept, put x=0. For the first line, y=b, while for the second line, y=6. Since both lines meet the y-axis at the same point, their y-intercepts must be equal. Hence, b=6. The number 5 is the slope of the first line, not its y-intercept. Exam tip: always set x=0 to find the y-intercept.