Slope and y-intercept

In the form \(y=mx+c\), \(m\) represents the slope. The slopes of the first and second lines are \(p+1\) and \(6\), respectively. For equal slopes, \(p+1=6\), so \(p=5\). If 6 were chosen, the first slope would be 7, not 6. Exam tip: identify the coefficient of \(x\) when comparing slopes.

In the form \(y=mx+c\), the constant term \(c\) is the \(y\)-intercept. The \(y\)-intercept of \(y=-4x+q\) is \(q\), while that of \(y=7x-2\) is \(-2\). Since the intercepts are equal, \(q=-2\). Note that \(-4\) is the slope of the first line, not its \(y\)-intercept. Exam tip: put \(x=0\) to find a \(y\)-intercept.

A line has the form y=mx+c, where c is its y-intercept. For the line to pass through the origin (0,0), its y-intercept must be 0. Hence, r+1=0, so r=-1. For example, if r=1, the y-intercept is 2, so that option is incorrect. Exam tip: In questions involving the origin, directly substitute x=0 and y=0.

Write the equation in slope-intercept form. In (y=(a-2)x+a), the slope is (a-2) and the y-intercept is a. If they were equal, then (a-2=a). Subtracting a from both sides gives (-2=0), which is impossible. Hence, no value of a satisfies the condition. Even for 0, 1, or 2, the slope remains 2 less than the y-intercept. Exam tip: in y=mx+c, the coefficient of x is the slope and the constant term is the y-intercept.

Substitute the point \((1,5)\) into \(y=2x+b\): \(5=2(1)+b\). Hence, \(b=5-2=3\). To verify, when \(x=3\), \(y=2(3)+3=9\), so the second point \((3,9)\) is also satisfied. Here, \(2\) is the slope, whereas \(b\) is the y-intercept. Exam tip: Substitute the coordinates of a given point into the line equation to find an unknown constant.

In \(x=-4\), the x-coordinate is fixed, so the graph is a vertical line parallel to the y-axis. It cannot meet the y-axis, where \(x=0\). In contrast, \(y=-4\) is horizontal. Exam tip: \(x=a\) always represents a vertical line.

In the form \(y=mx+c\), the constant term \(c\) is the \(y\)-intercept. The first line has \(y\)-intercept \(-4\), while the second has \(k-1\). For equal intercepts, \(k-1=-4\), so \(k=-3\). If \(-2\) were used, the second intercept would be \(-3\), not \(-4\). Exam tip: set \(x=0\) to identify a \(y\)-intercept.

In the form \(y=mx+c\), the coefficient of \(x\) is the slope. The slopes of the first and second lines are \(m-1\) and \(2\), respectively. Equal slopes give \(m-1=2\), so \(m=3\). If 2 were chosen, the first slope would be 1, not equal to the second slope. Exam tip: identify the coefficient of \(x\) directly to find the slope.

In \(y=mx+c\), \(m\) is the slope and \(c\) is the y-intercept. In option A, \(m=-2\) is negative while \(c=5\) is positive. Option C has a negative slope too, but its intercept is negative. Exam tip: check the signs of the x-coefficient and constant separately.

The slope is \(\frac{5}{2}\), meaning that an increase of 2 units in \(x\) increases \(y\) by 5 units. Hence, \(\Delta y=\frac{5}{2}\Delta x\). Putting \(\Delta y=20\), we get \(20=\frac{5}{2}\Delta x\), so \(\Delta x=20\times\frac{2}{5}=8\). If \(x\) increased by 10, \(y\) would increase by 25, not 20. 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 and q is the y-intercept. Since p=-4 is negative, y decreases as x increases, so the line decreases. Since q=9 is positive, the line cuts the y-axis above the origin. Option C is incorrect because a horizontal line must have slope 0. Exam tip: use the sign of p to identify whether the line rises or falls, and the sign of q to locate the y-intercept.

In \(x=-3\), the x-coordinate stays fixed, so the graph is a vertical line parallel to the y-axis. It cannot meet the y-axis, where \(x=0\). In contrast, \(x=0\) is the y-axis itself. Exam tip: a constant-x line has undefined slope.

Writing the given line as \(y=-\frac{3}{2}x+4\) gives slope \(-\frac{3}{2}\) and y-intercept 4. A perpendicular line has slope \(\frac{2}{3}\), so \(y=\frac{2}{3}x+4\), which is option C. Option B has intercept 4 but slope \(-\frac{2}{3}\). Exam tip: perpendicular slopes multiply to \(-1\).

A line in the form y=mx+c has y-intercept c. Here, 2n+1=11, so 2n=10 and n=5. Therefore, the slope is n-5=5-5=0. Option 1 cannot be the slope because substituting n=5 makes the coefficient of x zero. Exam tip: In y=mx+c, the coefficient of x is always the slope.

A line in the form \(y=mx+c\) has slope \(m\) and y-intercept \(c\). Here, the slope is \(r+4\). Thus, \(r+4=1\), giving \(r=-3\). Therefore, the y-intercept is \(r-2=-3-2=-5\). Note that \(-3\) is the value of \(r\), not the y-intercept. Exam tip: first find the parameter from the slope, then substitute it into the constant term.

An increase of \(10\) in \(y\) for an increase of \(2\) in \(x\) gives the slope \(m=\frac{10}{2}=5\). Since the line crosses the \(y\)-axis at \((0,-6)\), its \(y\)-intercept is \(b=-6\). Using \(y=mx+b\), the equation is \(y=5x-6\). The option \(y=10x-6\) has the correct intercept but an incorrect slope of \(10\). Exam tip: find the slope from the rate of change first, then identify the intercept by putting \(x=0\).

When \(x\) increases by 3 and \(y\) decreases by 12, the slope is \(m=\frac{-12}{3}=-4\). Since the line cuts the \(y\)-axis at \((0,9)\), its \(y\)-intercept is 9. Therefore, its equation is \(y=-4x+9\). Option A has a positive slope, but \(y\) is decreasing here. Exam tip: Use a negative sign for the slope whenever the dependent variable decreases.

In \(y=mx+c\), \(m=-\frac{2}{3}\) is negative, so the line falls from left to right. Since \(c=4\), its y-intercept is \((0,4)\). Exam tip: use \(m\) for direction and \(c\) for the y-intercept.

In the form y = mx + c, the y-intercept is the constant term c. Here, the y-intercept is t + 4. From t + 4 = 10, we get t = 6. Therefore, the slope is 2t - 3 = 2(6) - 3 = 9. Option 6 is the value of t, not the slope. Exam tip: set x = 0 to identify the y-intercept.

A straight line is commonly written as y = mx + c, where m is the slope and c is the y-intercept. Starting with 2x + 5y − 25 = 0, transpose the x-term and constant: 5y = −2x + 25. Divide every term by 5 to obtain y = (−2/5)x + 5. Therefore, the coefficient of x is the slope, −2/5, and the constant term is the y-intercept, 5, so option A is correct. Option B changes both signs, while options C and D read coefficients from the original equation without first isolating y.