Graphing a line - GMAT Quantitative

Let and be the - and -intercepts, respectively, of the line. Then the slope of the line is , or, equilvalently, .

We do not need to find the actual slopes of the four choices if we observe that in each case, and are of the same sign. Since the quotient of two numbers of the same sign is positive, it follows that is negative, and therefore, none of the pairs of intercepts can be those of a line with positive slope .

Let and be the - and -intercepts, respectively, of the line. Then the slope of the line is , or, equilvalently, .

We can examine the intercepts in each choice to determine which set meets these conditions.

and

Slope:

and

Slope:

and

Slope:

and

Slope:

and comprise the correct choice, since a line passing through these points has the correct slope.

Let and be the - and -intercepts, respectively, of the line. Then the slope of the line is , or, equilvalently, .

We can examine the intercepts in each choice to determine which set meets these conditions.

and :

Slope:

and

Slope:

and

Slope:

and

Slope:

and comprise the correct choice.

The - and -intercepts of the line are, respectively, and . If and are the - and -intercepts, respectively, of a line, the slope of the line is . This makes the slope of the green line .

Any line perpendicular to this line must have as its slope the opposite of the reciprocal of this, or . Since the desired line must also have -intercept , then the slope-intercept form of the line is

which can be rewritten in standard form:

The slope of the green line can be calculated by noting that the - and -intercepts of the line are, respectively, and . If and be the - and -intercepts, respectively, of a line, the slope of the line is . This makes the slope of the green line .

Any line perpendicular to this line must have as its slope the opposite reciprocal of this, or . Since the desired line must also have -intercept , the equation of the line, in point=slope form, is

which can be simplified as

If and be the - and -intercepts, respectively, of a line, the slope of the line is .

The - and -intercepts of the line are, respectively, and , so , and consequently, the slope of the green line is . A line parallel to this line must also have slope .

Each of the equations of the lines is in slope-intercept form , where is the slope, so we need only look at the coefficients of . The only choice that has as its -coefficient is , so this is the correct choice.

The -intercept of the line coincides with that of the graph of the quadratic equation, which is a parabola; to find the -intercept of the parabola, substitute 0 for in the quadratic equation:

The -intercept of the parabola, and of the line, is .

The -intercept of the line coincides with one of those of the parabola; to find the -intercepts of the parabola, substitute 0 for in the equation:

The quadratic expression can be "reverse-FOILed" by noting that 9 and have product and sum 7:

, in which case

or

, in which case .

The -intercepts of the parabola are and , so the -intercept of the line is one of these. We will examine both possibilities

If and be the - and -intercepts, respectively, of the line, then the slope of the line is . If the intercepts are and , the slope is ; if the intercepts are and , the slope is . Since the line is of positive slope, we choose the line of slope 9; since its -intercept is , then we can substitute in the slope-intercept form of the line, , to get the correct equation, .

The -intercept of the line coincides with that of the graph of the quadratic equation, which is a parabola; to find the -intercept of the parabola, substitute 0 for in the quadratic equation:

The -intercept of the parabola, and of the line, is .

The -intercept of the line coincides with one of those of the parabola; to find the -intercepts of the parabola, substitute 0 for in the equation:

Using the method, split the middle term by finding two integers whose product is and whose sum is ; by trial and error we find these to be and 4, so proceed as follows:

Split:

or

The -intercepts of the parabola are and , so the -intercept of the line is one of these. We examine both possibilities.

If and be the - and -intercepts, respectively, of the line, then the slope of the line is , or, equivalently,

If the intercepts are and , the slope is ; if the intercepts are and , the slope is . Since the line is of negative slope, we choose the line of slope ; since its -intercept is , then we can substitute in the slope-intercept form of the line, , to get the correct equation, .

The -intercept of the line coincides with that of the graph of the quadratic equation, which is a horizontal parabola; to find the -intercept of the parabola, substitute 0 for in the quadratic equation:

The -intercept of the parabola, and of the line, is .

The -intercept of the line coincides with one of those of the parabola; to find the -intercepts of the parabola, substitute 0 for in the equation:

Either

, in which case ,

or

, in which case .

The -intercepts of the parabola are and , so the -intercept of the line is one of these. We will examine both possibilities.

If and be the - and -intercepts, respectively, of the line, then the slope of the line is . If the intercepts of the line are and , the slope of the line is ; if the intercepts are and , the slope is . We choose the latter, since we are looking for a line with positive slope; since its -intercept is , then we can substitute in the slope-intercept form of the line, , to get the correct equation, .

The point of intersection of the two lines has as its coordinates the values of and that make both of the given linear equations true. Therefore, we seek to find the solution of the system of equations:

We need only find , so multiply both sides of the two equations by 7 and 4, respectively. Then add:

,

making 9 the correct response.

We can find out whether the graphs of and intersect by first solving for in the first equation:

We then substitute in the second equation for :

Then we rewrite in standard form:

Since we are only trying to pdetermine whether at least one point of intersection exists, rather than actually find the point, all we need to do is to evaluate the discriminant; if it is nonnegative, at least one solution - and, consequently, one point of intersection - exists. In the general quadratic equation , this is , so here, the discriminant is

.

Therefore, the line of the equation intersects the parabola of the equation .

We do the same for the other three lines:

Then we rewrite in standard form:

.

The line of intersects the parabola.

The line of intersects the parabola.

Since the discriminant is negative, the system has no real solution. This means that the line of does not intersect the parabola of the equation , and it is the correct choice.

To locate the -intercept of the equation , substitute 0 for :

The -intercept of the parabola is .

The vertex of the parabola of an equation of the form has -coordinate . Here, we substitute , to obtain -coordinate

.

To find the -coordinate, substitute this for :

The vertex is .

The line includes points and ; apply the slope formula:

The slope is , and the -intercept is ; in the slope-intercept form , substitute for and . The equation of the line is .

The vertex of the parabola of an equation of the form has -coordinate . Here, we substitute , to obtain -coordinate

.

To find the -coordinate, substitute this for :

The vertex is .

To find the -intercepts of the parabola, substitute 0 for in the equation:

Either

, in which case ,

or

, in which case .

The -intercepts are and .

The line includes and either or , so we find the slope in each case using the slope formula.

If the line includes and :

.

If the line includes and :

We choose the first case, since the line has positive slope. The line through and has as its equation, using the point-slope form with and point :

A line with undefined slope is a vertical line, and its equation is for some , so the -coordinate of all points it passes through is . If it goes through the vertex of a parabola , then the line has the equation . Therefore, all we need to find is the -coordinate of the vertex of the parabola.

The vertex of the parabola of the equation has as its -coordinate , which, for the parabola of the equation , can be found by setting :

The desired line is .

The intercepts of the line of the equation can be found by substituting 0 for each variable, in turn:

-intercept:

The -intercept is the point

-intercept:

The -intercept is the point .

This line and the axes together form a right triangle. The horizontal leg is the segment connecting the origin to , and its length is . The vertical leg is the segment connecting the origin to , and its length is . The area of a right triangle is half the product of these legs, which is

.

Of the five responses, 45 comes closest to the correct area.

The vertices of the triangle are the point of intersection of the graphs of the lines of the two equations, and the -intercepts of those lines.

The -intercept of the line of the equation can be found by setting and solving for :

The -intercept of the line of the equation can be found the same way:

The -intercepts are and ; these are two of the vertices of the triangle, and since the segment connecting them is horizontal, this will be taken as the base. The length of the base is the difference of the -coordinates:

The intersection of the lines of the equations and can be found by solving the system of linear equations, as follows:

The point of intersection, and the third vertex of the triangle, is .

Since we are taking the horiztonal segment to be the base, the height will be the vertical distance to this third point - namely, the -coordinate . The area is half the product of the base and the height:

Let and be the - and -intercepts, respectively, of the line. Then the slope of the line is , or, equilvalently, .

We do not need to find the actual slopes of the four choices if we observe that in each case, and are of the same sign. Since the quotient of two numbers of the same sign is positive, it follows that is negative, and therefore, none of the pairs of intercepts can be those of a line with positive slope .

Let and be the - and -intercepts, respectively, of the line. Then the slope of the line is , or, equilvalently, .

We can examine the intercepts in each choice to determine which set meets these conditions.

and

Slope:

and

Slope:

and

Slope:

and

Slope:

and comprise the correct choice, since a line passing through these points has the correct slope.

Let and be the - and -intercepts, respectively, of the line. Then the slope of the line is , or, equilvalently, .

We can examine the intercepts in each choice to determine which set meets these conditions.

and :

Slope:

and

Slope:

and

Slope:

and

Slope:

and comprise the correct choice.

The - and -intercepts of the line are, respectively, and . If and are the - and -intercepts, respectively, of a line, the slope of the line is . This makes the slope of the green line .

Any line perpendicular to this line must have as its slope the opposite of the reciprocal of this, or . Since the desired line must also have -intercept , then the slope-intercept form of the line is

which can be rewritten in standard form: