Slope and y-intercept

In \(x=3\), x is fixed, so the graph is vertical and parallel to the y-axis; therefore, its slope is undefined. Since it is not \(x=0\), it does not meet the y-axis. Exam tip: \(x=k\) represents a vertical line.

Rewriting option A gives \(y=-2x+4\). Therefore, its slope is \(-2\), which is negative, and its y-intercept is 4. Option D has a negative slope but intercepts the y-axis at 2. Exam tip: convert the equation to \(y=mx+c\) first.

The given line has slope \(-3\) and y-intercept \(7\). For perpendicular non-vertical lines, the product of slopes is \(-1\), so the required slope is \(\frac{1}{3}\). Keeping intercept \(7\) gives option A. Option B is parallel, not perpendicular. Exam tip: in \(y=mx+c\), \(c\) is the y-intercept.

For \(y=mx+c\), a line falls left to right when \(m<0\), and it meets the \(y\)-axis above the origin when \(c>0\). In option B, \(m=-4\) and \(c=7\). Option A has the correct intercept sign but a positive slope. Exam tip: check the signs of slope and intercept separately.

The slope is \(\frac{\Delta y}{\Delta x}=\frac{28}{4}=7\). The \(y\)-intercept is \(-10\), so in slope-intercept form \(y=mx+c\), substituting \(m=7\) and \(c=-10\) gives \(y=7x-10\). In \(y=28x-10\), the slope is 28, while in \(y=-7x-10\), \(y\) decreases as \(x\) increases. Exam tip: find slope by calculating \(\Delta y/\Delta x\).

When (x) increases by 5, (y) decreases by 30, so the slope is \(m=\frac{-30}{5}=-6\). Since the line passes through (0,14), its (y)-intercept is 14. Hence the equation is \(y=-6x+14\). In \(y=-30x+14\), the intercept is correct but the slope is -30, not -6. Exam tip: find the rate of change using \(\frac{\Delta y}{\Delta x}\).

The line is \,\(y=(a+4)x+(a-10)\). At the origin, \,\(x=0\) and \,\(y=0\), so \,\(a-10=0\), giving \,\(a=10\). The slope is the coefficient of \,\(x\): \,\(a+4=10+4=14\). Hence, 14 is correct. A value such as 12 would result from using an incorrect 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 c. Here, u-2=6, so u=8. Therefore, the slope is m=2u+1=2(8)+1=17. Option 16 would result from incorrectly omitting the +1 term. Exam tip: first use the y-intercept to find the parameter, then substitute it into the slope expression.

The governing concept is the slope-intercept form y = mx + c, where m is the slope and c is the y-intercept. Starting with 6x + 7y - 42 = 0, move the x-term and constant to the other side: 7y = -6x + 42. Divide every term by 7 to obtain y = (-6/7)x + 6. Therefore, the coefficient of x is the slope, -6/7, and the constant term is the y-intercept, 6. Option A is correct. Option B reverses both signs, while C and D incorrectly read coefficients from the original standard form without isolating y.

To find a y-intercept, put \(x=0\) in the equation of the line. The y-intercept of the first line is \(b\), while for the second line, putting \(x=0\) gives \(y=8\). Since both lines cut the y-axis at the same point, their y-intercepts must be equal: \(b=8\). The numbers \(6\) and \(-4\) are slopes, not y-intercepts. Exam tip: in \(y=mx+c\), \(c\) is the y-intercept directly.

Both lines have the same slope \(m\), so they have the same direction. Since \(c\ne d\), their y-intercepts differ; therefore, they are not coincident but distinct parallel lines. Exam tip: same slope with different y-intercepts means parallel lines.

A y-intercept of 4 means that y=4 when x=0, so c=4. Substituting x=-3 gives y=-9(-3)+4=27+4=31. The value 27 is only the product -9×(-3); the intercept 4 must also be added. Exam tip: in y=mx+c, the y-intercept is the constant term c.

The slope of the line is \(k-3\). Since \(y\) increases by \(48\) when \(x\) increases by \(6\), the slope is \(\frac{48}{6}=8\). Thus, \(k-3=8\), so \(k=11\). If \(k=10\), the slope would be only \(7\), giving a rise of \(42\) for a run of \(6\). Exam tip: in \(y=mx+c\), \(m\) is the slope.

The slope of the line is \(-(p-1)\). An increase of \(5\) in \(x\) and a decrease of \(35\) in \(y\) give \(\Delta y=-35\). Hence, the slope is \(\frac{-35}{5}=-7\). Therefore, \(-(p-1)=-7\), so \(p-1=7\) and \(p=8\). If \(p=7\), the slope would be \(-6\), not the required \(-7\). Exam tip: Write a decrease in \(y\) as a negative change before finding the slope.

In \(y=mx+c\), \(c\) is the \(y\)-intercept and \(m\) is the slope. Thus, the same intercept requires \(c=5\), while a different slope requires \(m\ne-2\). Option B gives the same slope. Exam tip: do not interchange the roles of \(m\) and \(c\).

In the form (y=mx+c), m represents the slope. Given m+c=15 and c=6, we get m+6=15. Therefore, m=15-6=9. Hence, the correct answer is 9. Option 8 is incorrect because subtracting 6 from 15 gives 9. Exam tip: In y=mx+c, identify m as the slope and c as the y-intercept.

After shifting upward by 4 units, the equation becomes \(y=mx+(c+4)\). Thus, \(c\), the y-intercept, changes, while \(m\) remains unchanged because the line’s steepness is unchanged. Exam tip: a vertical shift changes only the constant term.

Since the given point \((8,10)\) lies on the line, substitute \(x=8\) and \(y=10\) in \(y=\frac{3}{4}x+c\). This gives \(10=\frac{3}{4}\times8+c=6+c\), so \(c=4\). Hence, 4 is the correct option. If \(c=3\), the right-hand side would be 9, not the given y-coordinate 10. Exam tip: For a line passing through a point, directly substitute the point’s coordinates into its equation.

In \(y=mx+c\), \(m\) is the slope and \(c\) is the y-intercept. Here \(m=-\frac{3}{4}\) and \(c=5\); putting \(x=0\) gives \(y=5\). Option D has slope \(-\frac{4}{3}\), not \(-\frac{3}{4}\). Exam tip: a negative slope means the line falls from left to right.

In the form y = mx + c, the y-intercept is c. Here, a + 2 = 9, so a = 7. Therefore, the slope is 4a - 1 = 4(7) - 1 = 27. Getting 28 usually results from an error in the subtraction step. Exam tip: first compare the constant term with the y-intercept to find the parameter, then use the coefficient of x to find the slope.