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 - 4 and 8 respectively. For equal slopes, q - 4 = 8, so q = 12. Option 8 is the slope of the second line, not the value of q. Exam tip: identify the coefficient of x when finding a slope.

In the form \(y=mx+c\), \(c\) is the \(y\)-intercept. The \(y\)-intercept of \(y=-8x+s\) is \(s\), while that of \(y=5x-13\) is \(-13\). Since the intercepts are equal, \(s=-13\). Note that \(-8\) is the slope of the first line, not its intercept. Exam tip: put \(x=0\) to find the \(y\)-intercept.

Comparing the line with the standard form \(y=mx+c\), its slope is \(m=a+5\) and its \(y\)-intercept is \(c=a\). Therefore, their difference is \((a+5)-a=5\), which is true for every real value of a. Although it is true for \(a=5\) and \(a=0\), it is not restricted to those values. Exam tip: In \(y=mx+c\), the coefficient of x is the slope and the constant term is the \(y\)-intercept.

Substitute the point \((1,12)\) in \(y=5x+b\): \(12=5(1)+b\). Hence, \(b=12-5=7\). Checking with \(x=4\), we get \(y=5(4)+7=27\), so the second point also lies on the line. Taking \(b=5\) is incorrect because 5 is the slope, not the y-intercept. Exam tip: substitute the coordinates of a given point into the line equation to find an unknown constant.

In \(y=mx+c\), \(m=-2\) is negative, so the line falls from left to right. Since \(c=-3\), its y-intercept lies below the origin. Exam tip: check the signs of the slope and constant term separately.

In the form y = mx + c, the coefficient of x, m, is the slope. The slopes of the two lines are k - 7 and 4 respectively. Since the slopes are equal, k - 7 = 4, so k = 11. If k were 12, the first slope would be 5, not 4. Exam tip: when comparing slopes, equate only the coefficients of x.

In the form \(y=mx+c\), the y-intercept is the constant term \(c\). The first line has y-intercept \(-6\), while the second has y-intercept \(m-10\). For equal intercepts, \(m-10=-6\), so \(m=4\). If \(m=5\), the second intercept would be \(-5\), not \(-6\). Exam tip: put \(x=0\) to find a y-intercept.

\(m\) is the slope. If \(m>0\), the line rises from left to right; if \(m<0\), it falls. The y-intercept is determined by \(c\), not by the sign of \(m\). Exam tip: inspect the sign of \(m\) first.

The slope is \(\frac{5}{4}\), so an increase of 4 units in \(x\) increases \(y\) by 5 units. Thus, \(\Delta y=\frac{5}{4}\Delta x\). Putting \(\Delta y=25\), we get \(25=\frac{5}{4}\Delta x\), hence \(\Delta x=25\times\frac{4}{5}=20\). Option 25 is the required change in \(y\), not the change in \(x\). Exam tip: for change-based questions on a line, use the slope; the constant term \(-9\) does not affect the change.

The equation is in the form y = px + q, where p is the slope and q is the y-intercept. Since p = -7 is negative, the line goes downward as x increases, so it falls. Since q = 6 is positive, the line meets the y-axis at 6. Option C has the correct intercept but is wrong because a horizontal line must have slope 0. Exam tip: the sign of the slope tells whether a line rises or falls.

Both lines have the same slope, \(m\), so they are parallel. Since \(c_1\ne c_2\), their y-intercepts differ, so they cannot be the same line. Exam tip: equal slopes and different intercepts indicate distinct parallel lines.

Writing the given line as \(y=2x+5\) gives slope \(2\). A perpendicular line has slope \(-\frac12\). With y-intercept \(-3\), it is \(y=-\frac12x-3\), or \(x+2y+6=0\). Exam tip: perpendicular slopes multiply to \(-1\).

In the form \(y=mx+c\), the \(y\)-intercept is the constant term \(c\). Here, \(3n-2=19\), so \(3n=21\) and \(n=7\). Hence, the slope is \(n-6=7-6=1\). Option 2 could result from an incorrect calculation of \(n-6\). Exam tip: first use the constant term to find the parameter, then take the coefficient of \(x\) as the slope.

\(x=3\) is a vertical line parallel to the y-axis, \(x=0\), so it never meets the y-axis and has no y-intercept. In contrast, \(x+y=3\) gives \(y=3\) when \(x=0\). Exam tip: \(x=c\), for \(c\ne0\), has no y-intercept.

The slope is \(m=\frac{\Delta y}{\Delta x}=\frac{40}{5}=8\). Since the line meets the (y)-axis at (0,-11), its (y)-intercept is \(-11\). Substituting \(m=8\) and \(c=-11\) in \(y=mx+c\) gives \(y=8x-11\). The equation \(y=40x-11\) has slope 40, while \(y=-8x-11\) represents a decreasing line. Exam tip: calculate slope as \(\Delta y/\Delta x\).

As \(x\) increases by \(6\), \(y\) decreases by \(42\), so the slope is \(m=\frac{-42}{6}=-7\). The line crosses the \(y\)-axis at \((0,16)\), hence its \(y\)-intercept is \(16\). Therefore, the equation is \(y=-7x+16\). In \(y=-42x+16\), the slope is \(-42\), not the required rate of change. Exam tip: find the slope using \(\frac{\Delta y}{\Delta x}\) first, then use the \(y\)-intercept.

The equation of the line is \(y=(a+7)x+(a-12)\). Since it passes through the origin \((0,0)\), substitute \(x=0\) and \(y=0\) to get \(a-12=0\), so \(a=12\). Hence, the slope, the coefficient of \(x\), is \(a+7=12+7=19\). Option \(18\) is a close distractor, but it does not result from substituting the correct value of \(a\). Exam tip: in \(y=mx+c\), \(m\) is the slope and \(c\) is the y-intercept.

In the form y = mx + c, the y-intercept is the constant term c. Here, the y-intercept is u+5, so u+5=14 gives u=9. Therefore, the slope is 2u-3=2(9)-3=15. The value 13 is a distractor because it does not result from substituting the correct value of u into the slope expression. Exam tip: in y=mx+c, the coefficient of x is the slope and the constant term is the y-intercept.

From (3y=-8x+30), (y=-\frac{8}{3}x+10). So the (y)-intercept is (10) and slope is (-\frac{8}{3}).

To find a line’s y-intercept, put x=0. The y-intercept of y=7x+b is b, while that of y=-5x+9 is 9. Since both lines cut the y-axis at the same point, their y-intercepts must be equal: b=9. The value 12 may come from the difference of the slopes, but slopes do not determine the y-intercept. Exam tip: for a y-intercept question, always substitute x=0.