Slope and y-intercept

In the form \(y=mx+c\), the coefficient of \(x\) is the slope. Since the slopes are equal, \(a=6\). Substituting \(x=3\) in the first line gives \(y=6\times3-7=11\). The \(5\) is the y-intercept of the second line, so it does not affect the y-value of the first line. Exam tip: first equate the slopes to find the unknown coefficient, then substitute the given value of \(x\).

A y-intercept of 5 means that when x=0, y=5; therefore, c=5. The equation is y=-8x+5. Substituting x=-2 gives y=-8(-2)+5=16+5=21. Option 20 may seem close, but it does not result from adding the constant term 5 correctly. Exam tip: When substituting a negative value of x, check the sign of the product carefully.

The given line has slope \(2\). For perpendicular lines, the product of slopes is \(-1\), so the required slope is \(-\frac{1}{2}\). With y-intercept \(3\), the equation is \(y=-\frac{1}{2}x+3\). Option D has the correct slope but intercept \(-3\). Exam tip: check \(m\) and \(c\) separately in \(y=mx+c\).

The slope is \((420-260)/(7-3)=40\) rupees per km. Substituting \(x=3, y=260\) in \(y=40x+b\) gives \(b=140\). Option B wrongly treats the charge at 3 km as the y-intercept. Exam tip: put \(x=0\) to check the fixed charge.

\(x=4\) is a vertical line. Its change in \(x\) is zero, so the slope \(\Delta y/\Delta x\) is undefined. Exam tip: any line of the form \(x=\text{constant}\) has an undefined slope.

Given \(m+c=12\) and \(c=5\), substituting the value of \(c\) gives \(m+5=12\). Hence, \(m=7\). In the equation \(y=mx+c\), \(m\) represents the slope, so the correct answer is \(7\). The value \(5\) is the y-intercept \(c\), not the slope. Exam tip: In \(y=mx+c\), the coefficient of \(x\), namely \(m\), is always the slope.

In the form \(y=mx+c\), \(m\) is the slope and \(c\) is the \(y\)-intercept. Given slope \(m=-4\). Substituting in \(m+c=3\) gives \(-4+c=3\), so \(c=7\). Choosing \(5\) would give \(m+c=1\), which does not satisfy the condition. Exam tip: In \(y=mx+c\), identify \(c\) directly as the \(y\)-intercept.

Since the point \((6,9)\) lies on the line, substitute \(x=6\) and \(y=9\): \(9=\frac{2}{3}\times 6+c=4+c\). Hence, \(c=9-4=5\). Option B, 4, is only the value of \(\frac{2}{3}\times6\), not the y-intercept. Exam tip: To find an unknown constant in a line equation, substitute the coordinates of the given point.

The given line has slope \(\frac{1}{3}\). Slopes of perpendicular lines have product \(-1\), so the required slope is \(-3\). With y-intercept 4, the equation is \(y=-3x+4\). Exam tip: check the slope and constant term separately.

In the form \(y=mx+c\), the \(y\)-intercept is the constant term \(c\). Here, \(a+1=6\), so \(a=5\). Therefore, the slope is \(m=3a-2=3(5)-2=13\). Option 12 would result from incorrectly ignoring the \(-2\). Exam tip: first use the \(y\)-intercept to find the parameter, then substitute it into the coefficient of \(x\).

Writing \(2x+4y=8\) as \(y=-\frac12x+2\) gives slope \(-\frac12\). A perpendicular line has slope \(2\); with y-intercept \(-3\), it is \(y=2x-3\). Exam tip: perpendicular slopes multiply to \(-1\).

The governing idea is that in y = mx + c, m is the slope and c is the y-intercept. Rearrange the equation carefully: 2x − 3y + 9 = 0 gives −3y = −2x − 9. Dividing each term by −3 produces y = (2/3)x + 3. Consequently, the slope is 2/3 and the line crosses the y-axis at (0, 3), so option A is correct. The negative signs in option B would correspond to an incorrect division. Options C and D do not result from isolating y and therefore mix up coefficients from the general form. The y-intercept can independently be checked by setting x = 0.

Both equations are already in the form y = mx + c, so their slopes are the coefficients of x. Equal slopes give t + 1 = 8, which implies t = 7. The second line has constant term 2t − 3, so its y-intercept is 2(7) − 3 = 14 − 3 = 11. Therefore option A is correct. The given option list has 11 as the first choice; the earlier marked choice C was inconsistent with the calculation and has been corrected. The value −4 belongs to the first line, while 8 is a slope, not an intercept. Substituting t = 7 gives the second equation y = 8x + 11, confirming the result.

The y-intercept of the first line is \(v-4\), while that of the second line is \(5\). Since the y-intercepts are equal, \(v-4=5\), so \(v=9\). The slope of the second line \(y=(v+1)x+5\) is \(v+1=9+1=10\). Note that \(9\) is the value of \(v\), not the slope. Exam tip: in \(y=mx+c\), \(m\) is the slope and \(c\) is the y-intercept.

The slope is \(-\frac{3}{2}\). Thus, for every increase of 2 units in \(x\), \(y\) decreases by 3 units. If the increase in \(x\) is \(\Delta x\), the decrease in \(y\) is \(\frac{3}{2}\Delta x\). Hence, \(\frac{3}{2}\Delta x=18\), so \(\Delta x=12\). If 13 were chosen, the decrease would be \(\frac{3}{2}\times13=19.5\), not 18. Exam tip: With a negative slope, increasing \(x\) decreases \(y\).

The slope is
\(m=(5-(-3))/(4-0)=2\). A y-intercept is the y-value when
\(x=0\), so it is -3 here. The value 5 is only the y-coordinate of
(4,5). Exam tip: first locate the point with
\(x=0\).

Given \(c=-6\) and \(m-c=14\), substitute the value of \(c\): \(m-(-6)=14\), so \(m+6=14\). Hence, \(m=8\). Therefore, the slope of the line is 8. Choosing 6 is a common error; it is the magnitude of the negative value of \(c\), not the slope. Exam tip: always use brackets when substituting a negative value in \(m-c\).

Given m=-2 and 2c+m=10, substituting m gives 2c-2=10. Hence, 2c=12 and c=6. In the line y=mx+c, the y-intercept is c, so the correct answer is 6. Choosing 5 is incorrect because it does not satisfy 2c-2=10. Exam tip: In y=mx+c, identify the constant term c to find the y-intercept.

In \(y=mx+c\), \(m\) is the slope and \(c\) is the y-intercept. In option A, \(m=2\) is positive and \(c=-3\) is negative. Option B has the correct intercept sign but a negative slope. Exam tip: check the signs of the coefficient of \(x\) and the constant separately.

In \(x=5\), x remains fixed, so the line is vertical, parallel to the y-axis, and has an undefined slope. \(y=5\) is horizontal with slope 0. Exam tip: \(x=\text{constant}\) represents a vertical line.